ICSE Class 10 Mathematics · RS Aggarwal Mathematics · 83 solved questions
Using remainder theorem, find the remainder when:
f(x) = 3x2 - 5x + 7 is divided by (x - 2).
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
f(x) = 3x2 - 5x + 7
Divisor :
⇒ x - 2 = 0
⇒ x = 2
Substituting x = 2 in f(x), we get :
⇒ f(2) = 3(2)2 - 5(2) + 7
= 3(4) - 5(2) + 7
= 12 - 10 + 7
= 9.
Hence, remainder = 9.
Using remainder theorem, find the remainder when:
f(x) = 2x3 - 5x2 + 3x - 10 is divided by (x - 3).
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
f(x) = 2x3 - 5x2 + 3x - 10
Divisor :
⇒ x - 3 = 0
⇒ x = 3
Substituting x = 3 in f(x), we get :
⇒ f(3) = 2(3)3 - 5(3)2 + 3(3) - 10
= 2(27) - 5(9) + 3(3) - 10
= 54 - 45 + 9 - 10
= 8.
Hence, remainder = 8.
Using remainder theorem, find the remainder when:
f(x) = 5x3 - 12x2 + 17x - 6 is divided by (x - 1).
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
Let f(x) = 5x3 - 12x2 + 17x - 6
Divisor :
⇒ x - 1 = 0
⇒ x = 1.
Substituting x = 1 in f(x), we get :
⇒ f(1) = 5(1)3 - 12(1)2 + 17(1) - 6
= 5(1) - 12(1) + 17(1) - 6
= 5 - 12 + 17 - 6
= 4.
Hence, remainder = 4.
Using remainder theorem, find the remainder when:
f(x) = x3 - 2x2 - 5x + 6 is divided by (x + 2).
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
Let f(x) = x3 - 2x2 - 5x + 6
Divisor :
⇒ x + 2 = 0
⇒ x = -2
Substituting x = -2 in f(x), we get :
⇒ f(-2) = (-2)3 - 2(-2)2 - 5(-2) + 6
= (-8) - 2(4) - 5(-2) + 6
= -8 - 8 + 10 + 6
= 0.
Hence, remainder = 0.
Using remainder theorem, find the remainder when:
f(x) = 8x3 - 16x2 + 14x - 5 is divided by (2x - 1).
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
Let f(x) = 8x3 - 16x2 + 14x - 5
Divisor :
⇒ (2x - 1) = 0
⇒ 2x = 1
⇒ x = $\frac{1}{2}$
Substituting x = $\frac{1}{2}$ in f(x), we get :
$$\begin{aligned}\Rightarrow f\Big(\frac{1}{2}\Big) = 8\Big(\frac{1}{2}\Big)^3 - 16\Big(\frac{1}{2}\Big)^2 + 14\Big(\frac{1}{2}\Big) - 5 \\[1em] = 8\Big(\frac{1}{8}\Big) - 16\Big(\frac{1}{4}\Big) + 14\Big(\frac{1}{2}\Big) - 5 \\[1em] = 1 - 4 + 7 - 5 \\[1em] = -1.\end{aligned}$$
Hence, remainder = -1.
Using remainder theorem, find the remainder when:
f(x) = 9x2 - 6x + 2 is divided by (3x - 2).
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
f(x) = 9x2 - 6x + 2
Divisor :
⇒ (3x - 2) = 0
⇒ 3x = 2
⇒ x = $\frac{2}{3}$
Substituting x = $\frac{2}{3}$ in f(x), we get :
$$\begin{aligned}\Rightarrow f\Big(\frac{2}{3}\Big) = 9\Big(\frac{2}{3}\Big)^2 - 6\Big(\frac{2}{3}\Big) + 2 \\[1em] = 9\Big(\frac{4}{9}\Big) - 6\Big(\frac{2}{3}\Big) + 2 \\[1em] = 4 - 4 + 2 \\[1em] = 2.\end{aligned}$$
Hence, remainder = 2.
Using remainder theorem, find the remainder when:
f(x) = 8x2 - 2x - 15 is divided by (2x + 3).
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
f(x) = 8x2 - 2x - 15
Divisor :
⇒ 2x + 3 = 0
⇒ 2x = -3
⇒ x = -$\frac{3}{2}$
Substituting x = -$\frac{3}{2}$ in f(x), we get :
$$\begin{aligned}\Rightarrow f\Big(-\frac{3}{2}\Big) = 8\Big(-\frac{3}{2}\Big)^2 - 2\Big(-\frac{3}{2}\Big) - 15 \\[1em] = 8\Big(\frac{9}{4}\Big) + 2\Big(\frac{3}{2}\Big) - 15 \\[1em] = 18 + 3 - 15 \\[1em] = 6.\end{aligned}$$
Hence, remainder = 6.
On dividing (ax3 + 9x2 + 4x - 10) by (x + 3), we get 5 as remainder. Find the value of a.
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
Let f(x) = ax3 + 9x2 + 4x - 10
Given,
Remainder = 5
Divisor :
⇒ x + 3 = 0
⇒ x = -3.
Substituting x = -3 in f(x), will give remainder 5.
⇒ f(-3) = 5
⇒ a(-3)3 + 9(-3)2 + 4(-3) - 10 = 5
⇒ a(-27) + 81 - 12 - 10 = 5
⇒ -27a + 81 - 22 = 5
⇒ -27a + 59 = 5
⇒ -27a = 5 - 59
⇒ -27a = -54
⇒ a = $\frac{-54}{-27}$
⇒ a = 2.
Hence, the value of a = 2.
Using Remainder Theorem, find the value of k if on dividing 2x3 + 3x2 - kx + 5 by (x - 2), leaves a remainder 7.
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
Let f(x) = 2x3 + 3x2 - kx + 5
Given,
Remainder = 7
Divisor :
⇒ x - 2 = 0
⇒ x = 2
Substituting x = 2 in f(x), gives remainder 7.
⇒ f(2) = 7
⇒ 2(2)3 + 3(2)2 - k(2) + 5 = 7
⇒ 2(8) + 3(4) - 2k + 5 = 7
⇒ 16 + 12 - 2k + 5 = 7
⇒ -2k + 33 = 7
⇒ 2k = 33 - 7
⇒ 2k = 26
⇒ k = $\frac{26}{2}$
⇒ k = 13.
Hence, the value of k is 13.
If the polynomials 2x3 + ax2 + 3x - 5 and x3 + x2 - 2x + a leave the same remainder when divided by (x - 2), find the value of a. Also, find the remainder in each case.
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
Let, p(x) = 2x3 + ax2 + 3x - 5 and q(x) = x3 + x2 - 2x + a
Divisor :
⇒ x - 2
⇒ x = 2
⇒ p(2) = 2(2)3 + a(2)2 + 3(2) - 5
= 2(8) + 4a + 6 - 5
= 16 + 4a + 1
= 4a + 17.
⇒ q(2) = (2)3 + (2)2 - 2(2) + a
= 8 + 4 - 4 + a
= 8 + a.
Given,
Polynomials 2x3 + ax2 + 3x - 5 and x3 + x2 - 2x + a leave the same remainder when divided by (x - 2).
∴ p(2) = q(2)
⇒ 4a + 17 = 8 + a
⇒ 4a - a = 8 - 17
⇒ 3a = -9
⇒ a = $\frac{-9}{3}$
⇒ a = -3.
Substituting value of a in p(2) :
⇒ p(2) = 4a + 17
= 4(-3) + 17
= -12 + 17
= 5.
Substituting value of a in q(2) :
⇒ q(2) = 8 + a
= 8 - 3
= 5.
Hence, the value of a = -3 and remainder in each case is 5.
The polynomials f(x) = ax3 + 3x2 - 3 and g(x) = 2x3 - 5x + a when divided by (x - 4) leave the same remainder in each case. Find the value of a.
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
Given,
f(x) = ax3 + 3x2 - 3
g(x) = 2x3 - 5x + a
Divisor :
⇒ x - 4 = 0
⇒ x = 4
On dividing ax3 + 3x2 - 3 by x - 4,
⇒ f(4) = a(4)3 + 3(4)2 - 3
= 64a + 48 - 3
= 64a + 45.
On dividing 2x3 - 5x + a by x - 4,
⇒ g(4) = 2(4)3 - 5(4) + a
= 128 - 20 + a
= 108 + a.
Given,
On dividing by (x - 4) polynomials f(x) = ax3 + 3x2 - 3 and g(x) = 2x3 - 5x + a leave same remainder.
⇒ f(4) = g(4)
⇒ 64a + 45 = 108 + a
⇒ 64a - a = 108 - 45
⇒ 63a = 63
⇒ a = $\frac{63}{63}$
⇒ a = 1.
Hence, the value of a = 1.
Find a if the two polynomials ax3 + 3x2 - 9 and 2x3 + 4x + a leave the same remainder when divided by (x + 3).
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
Let p(x) = ax3 + 3x2 - 9 and q(x) = 2x3 + 4x + a
Given,
Divisor :
⇒ x + 3 = 0
⇒ x = -3
On dividing ax3 + 3x2 - 9 by x + 3, we get :
⇒ p(-3) = a(-3)3 + 3(-3)2 - 9
= -27a + 27 - 9
= -27a + 18.
On dividing 2x3 + 4x + a by x + 3, we get :
⇒ q(-3) = 2(-3)3 + 4(-3) + a
= -54 - 12 + a
= -66 + a.
Given,
Polynomials ax3 + 3x2 - 9 and 2x3 + 4x + a leave the same remainder when divided by (x + 3).
∴ p(-3) = q(-3)
⇒ -27a + 18 = -66 + a
⇒ -27a - a = -66 - 18
⇒ -28a = -84
⇒ a = $\frac{-84}{-28}$
⇒ a = 3.
Hence, the value of a = 3.
If (2x3 + ax2 + bx - 2) when divided by (2x - 3) and (x + 3) leaves remainders 7 and -20 respectively, find values of a and b.
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
Let, f(x) = 2x3 + ax2 + bx - 2.
Given,
Divisor :
⇒ 2x - 3 = 0
⇒ 2x = 3
⇒ x = $\frac{3}{2}$
Given,
On dividing 2x3 + ax2 + bx - 2 by 2x - 3, remainder is 7.
$$\begin{aligned}\Rightarrow f\Big(\frac{3}{2}\Big) = 7 \\[1em] \Rightarrow 2\Big(\frac{3}{2}\Big)^3 + a\Big(\frac{3}{2}\Big)^2 + b\Big(\frac{3}{2}\Big) - 2 = 7 \\[1em] \Rightarrow 2\Big(\frac{27}{8}\Big) + \Big(\frac{9a}{4}\Big) + \Big(\frac{3b}{2}\Big) - 2 = 7 \\[1em] \Rightarrow \frac{27}{4} + \frac{9a}{4} + \frac{3b \times 2}{4} = 7 + 2 \\[1em] \Rightarrow \Big(\frac{27 + 9a + 6b}{4}\Big) = 9 \\[1em] \Rightarrow 27 + 9a + 6b = 9 \times 4 \\[1em] \Rightarrow 27 + 9a + 6b = 36 \\[1em] \Rightarrow 9a + 6b = 36 - 27 \\[1em] \Rightarrow 9a + 6b = 9 \\[1em] \Rightarrow 3(3a + 2b) = 9 \\[1em] \Rightarrow 3a + 2b = \frac{9}{3} \\[1em] \Rightarrow 3a + 2b = 3.....(1)\end{aligned}$$
Divisor :
⇒ x + 3 = 0
⇒ x = -3
On dividing 2x3 + ax2 + bx - 2 by x + 3, remainder is -20.
⇒ f(-3) = -20
⇒ 2(-3)3 + a(-3)2 + b(-3) - 2 = -20
⇒ 2(-27) + 9a - 3b - 2 = -20
⇒ -54 + 9a - 3b - 2 = -20
⇒ 9a - 3b - 56 = -20
⇒ 9a - 3b = -20 + 56
⇒ 9a - 3b = 36
⇒ 3(3a - b) = 36
⇒ 3a - b = $\frac{36}{3}$
⇒ 3a - b = 12
⇒ b = 3a - 12 ....(2)
Substituting value of b from equation (2) in 3a + 2b = 3, we get :
⇒ 3a + 2(3a - 12) = 3
⇒ 3a + 6a - 24 = 3
⇒ 9a = 27
⇒ a = $\frac{27}{9}$
⇒ a = 3.
Substituting value of a in equation (2), we get :
⇒ b = 3(3) - 12
⇒ b = 9 - 12
⇒ b = -3.
Hence, the value of a = 3 and b = -3.
Using the Remainder Theorem, find the remainders obtained when x3 + (kx + 8)x + k is divided by x + 1 and x - 2. Hence find k if the sum of the two remainders is 1.
By remainder theorem,
If f(x) is divided by (x - a), then remainder = f(a).
Let f(x) = x3 + (kx + 8)x + k = x3 + kx2 + 8x + k
Given,
Divisor :
⇒ x + 1 = 0
⇒ x = -1
On dividing x3 + kx2 + 8x + k by x + 1, we get :
⇒ f(-1) = (-1)3 + k(-1)2 + 8(-1) + k
= -1 + k - 8 + k
= 2k - 9.
Divisor :
⇒ x - 2 = 0
⇒ x = 2.
On dividing x3 + kx2 + 8x + k by x - 2, we get :
⇒ f(2) = (2)3 + k(2)2 + 8(2) + k
= 8 + 4k + 16 + k
= 5k + 24
Given,
Sum of two remainders is 1.
⇒ 2k - 9 + 5k + 24 = 1
⇒ 7k + 15 = 1
⇒ 7k = 1 - 15
⇒ k = $-\frac{14}{7}$
⇒ k = -2.
Hence, the value of k = -2.
Using factor theorem, show that:
(x - 3) is a factor of (x3 + x2 - 17x + 15).
Let f(x) = (x3 + x2 - 17x + 15).
Given,
Divisor :
⇒ x - 3 = 0
⇒ x = 3
By factor theorem,
(x - a) is a factor of f(x), if f(a) = 0.
Substituting x = 3 in f(x), we get :
⇒ f(3) = (3)3 + (3)2 - 17(3) + 15
= 27 + 9 - 51 + 15
= 51 - 51
= 0.
Since, f(3) = 0, thus (x - 3) is a factor of f(x).
Hence, proved that (x - 3) is factor of x3 + x2 - 17x + 15.
Using factor theorem, show that:
(x + 1) is a factor of (x3 + 4x2 + 5x + 2).
Let f(x) = (x3 + 4x2 + 5x + 2)
Given,
Divisor:
⇒ x + 1 = 0
⇒ x = -1
By factor theorem,
(x - a) is a factor of f(x), if f(a) = 0.
Substituting x = -1 in f(x), we get :
⇒ f(-1) = (-1)3 + 4(-1)2 + 5(-1) + 2
= -1 + 4(1) - 5 + 2
= -1 + 4 - 5 + 2
= 6 - 6
= 0.
Since, f(-1) = 0, thus (x + 1) is a factor of f(x).
Hence, proved that (x + 1) is factor of x3 + 4x2 + 5x + 2.
Using factor theorem, show that:
(3x - 2) is a factor of (3x3 + x2 - 20x + 12).
Let f(x) = (3x3 + x2 - 20x + 12)
Given,
Divisor:
⇒ 3x - 2 = 0
⇒ 3x = 2
⇒ x = $\frac{2}{3}$
By factor theorem,
(x - a) is a factor of f(x), if f(a) = 0.
Substituting x = $\Big(\frac{2}{3}\Big)$ in f(x), we get :
$$\begin{aligned}\Rightarrow f\Big(\frac{2}{3}\Big) = 3\Big(\frac{2}{3}\Big)^3 + \Big(\frac{2}{3}\Big)^2 - 20\Big(\frac{2}{3}\Big) + 12 \\[1em] = \Big(\frac{8}{9}\Big) + \Big(\frac{4}{9}\Big) - \Big(\frac{40}{3}\Big) + 12 \\[1em] = \Big(\frac{8 + 4 - 120 + 108}{9}\Big) \\[1em] = \Big(\frac{120 - 120}{9}\Big) \\[1em] = 0.\end{aligned}$$
Since, $f\Big(\frac{2}{3}\Big)$ = 0, thus (3x - 2) is a factor of f(x).
Hence, proved that (3x - 2) is factor of 3x3 + x2 - 20x + 12.
Using factor theorem, show that:
(3 - 2x) is a factor of (2x3 - 9x2 + x + 12).
Let f(x) = (2x3 - 9x2 + x + 12)
Given,
Divisor:
⇒ 3 - 2x = 0
⇒ 2x = 3
⇒ x = $\frac{3}{2}$
By factor theorem,
(x - a) is a factor of f(x), if f(a) = 0.
Substituting x = $\Big(\frac{3}{2}\Big)$ in f(x), we get :
$$\begin{aligned}\Rightarrow f\Big(\frac{3}{2}\Big) = 2\Big(\frac{3}{2}\Big)^3 - 9\Big(\frac{3}{2}\Big)^2 + \Big(\frac{3}{2}\Big) + 12 \\[1em] = 2\Big(\frac{27}{8}\Big) - 9\Big(\frac{9}{4}\Big) + \Big(\frac{3}{2}\Big) + 12 \\[1em] = \frac{27}{4} - \frac{81}{4} + \frac{3}{2} + 12 \\[1em] = \Big(\frac{27 - 81 + 6 + 48}{4}\Big) \\[1em] = \frac{81 - 81}{27} \\[1em] = 0.\end{aligned}$$
Since, f$\Big(\frac{3}{2}\Big)$ = 0, thus (3 - 2x) is a factor of f(x).
Hence, proved that (3 - 2x) is factor of 2x3 - 9x2 + x + 12.
Use factor theorem to show that (x + 2) and (2x - 3) are factors of (2x2 + x - 6).
Let f(x) = 2x2 + x - 6
Given,
Factors : (x + 2) and (2x - 3)
⇒ x + 2 = 0 and 2x - 3 = 0
⇒ x = -2 and 2x = 3
⇒ x = -2 and x = $\frac{3}{2}$
By factor theorem,
(x - a) is a factor of f(x), if f(a) = 0.
Thus, (x + 2) and (2x - 3) are factors of f(x), if f(-2) = 0 and f$\Big(\frac{3}{2}\Big)$ = 0.
On dividing 2x2 + x - 6 by (x + 2), we get :
⇒ f(-2) = 2(-2)2 + (-2) - 6
= 2(4) - 2 - 6
= 8 - 2 - 6
= 8 - 8
= 0.
On dividing 2x2 + x - 6 by (2x - 3), we get :
$$\begin{aligned}\Rightarrow f\Big(\frac{3}{2}\Big) = 2\Big(\frac{3}{2}\Big)^2 + \Big(\frac{3}{2}\Big) - 6 \\[1em] = 2\Big(\frac{9}{4}\Big) + \Big(\frac{3}{2}\Big) - 6 \\[1em] = \Big(\frac{9}{2}\Big) + \Big(\frac{3}{2}\Big) - 6 \\[1em] = \Big(\frac{9 + 3}{2}\Big) - 6 \\[1em] = \Big(\frac{12}{2}\Big) - 6 \\[1em] = 6 - 6 \\[1em] = 0.\end{aligned}$$
Since, f(-2) = $f\Big(\frac{3}{2}\Big)$ = 0.
Hence, proved that x + 2 and 2x - 3 are factors of 2x2 + x - 6.
Find the value of a so that (x + 6) is a factor of the polynomial (x3 + 5x2 - 4x + a).
Let f(x) = x3 + 5x2 - 4x + a.
Given,
Factor: x + 6
Thus, on dividing x3 + 5x2 - 4x + a by x + 6, remainder will be zero.
⇒ f(-6) = 0
⇒ (-6)3 + 5(-6)2 - 4(-6) + a = 0
⇒ -216 + 180 + 24 + a = 0
⇒ -12 + a = 0
⇒ a = 12.
Hence, the value of a = 12.
For what value of a is the polynomial (2x3 + ax2 + 11x + a + 3) exactly divisible by (2x - 1)?
Let f(x) = 2x3 + ax2 + 11x + a + 3.
Given,
Factor : 2x - 1
⇒ 2x - 1 = 0
⇒ 2x = 1
⇒ x = $\frac{1}{2}$
Thus, on dividing 2x3 + ax2 + 11x + a + 3 by 2x - 1, remainder = 0.
$$\begin{aligned}\therefore f\Big(\frac{1}{2}\Big) = 0 \\[1em] \Rightarrow 2\Big(\frac{1}{2}\Big)^3 + a\Big(\frac{1}{2}\Big)^2 + 11\Big(\frac{1}{2}\Big) + a + 3 = 0 \\[1em] \Rightarrow 2\Big(\frac{1}{8}\Big) + \Big(\frac{a}{4}\Big) + \Big(\frac{11}{2}\Big) + a + 3 = 0 \\[1em] \Rightarrow \Big(\frac{1}{4}\Big) + \Big(\frac{a}{4}\Big) + \Big(\frac{11}{2}\Big) + a + 3 = 0 \\[1em] \Rightarrow \Big(\frac{1 + a + 22 + 4a + 12}{4}\Big) = 0 \\[1em] \Rightarrow 1 + a + 22 + 4a + 12 = 0 \\[1em] \Rightarrow 5a + 35 = 0 \\[1em] \Rightarrow 5a = -35 \\[1em] \Rightarrow a = \frac{-35}{5} = -7.\end{aligned}$$
Hence, the value of a = -7.
What must be subtracted from the polynomial x3 + x2 - 2x + 1, so that the result is exactly divisible by (x - 3)?
Polynomial : x3 + x2 - 2x + 1
Division by x - 3
⇒ x - 3 = 0
⇒ x = 3.
Let k be subtracted from the polynomial, so resulting polynomial is x3 + x2 - 2x + 1 - k.
Resulting polynomial should be exactly divisible by x - 3,
Thus, substituting x = 3, in polynomial x3 + x2 - 2x + 1 - k, remainder = 0.
⇒ 33 + 32 - 2(3) + 1 - k = 0
⇒ 27 + 9 - 6 + 1 - k = 0
⇒ 31 - k = 0
⇒ k = 31.
Hence, k = 31.
What must be subtracted from 16x3 - 8x2 + 4x + 7 so that the resulting expression has (2x + 1) as a factor?
Let the number to be subtracted from 16x3 - 8x2 + 4x + 7 be a.
Resulting polynomial [f(x)] = 16x3 - 8x2 + 4x + 7 - a
Given,
Factor: 2x + 1
⇒ 2x + 1 = 0
⇒ 2x = -1
⇒ x = $-\frac{1}{2}$.
Since, 2x + 1 is a factor.
Thus, on dividing 16x3 - 8x2 + 4x + 7 - a by 2x + 1, remainder = 0.
$$\begin{aligned}\therefore f\Big(\frac{-1}{2}\Big) = 0 \\[1em] \Rightarrow 16\Big(\frac{-1}{2}\Big)^3 - 8\Big(\frac{-1}{2}\Big)^2 + 4\Big(\frac{-1}{2}\Big) + 7 - a = 0 \\[1em] \Rightarrow 16\Big(\frac{-1}{8}\Big) - 8\Big(\frac{1}{4}\Big) - 2 + 7 - a = 0 \\[1em] \Rightarrow -2 - 2 - 2 + 7 - a = 0 \\[1em] \Rightarrow -6 + 7 - a = 0 \\[1em] \Rightarrow a = 1.\end{aligned}$$
Hence, the required number to be subtracted from the polynomial = 1.
Using factor theorem, show that (x - 3) is a factor of (x3 - 7x2 + 15x - 9). Hence, factorize the given expression completely.
Let f(x) = x3 - 7x2 + 15x - 9.
We know that,
(x - 3) will be the factor of f(x), if f(3) will be equal to 0.
⇒ f(3) = (3)3 - 7(3)2 + 15(3) - 9.
= 27 - 63 + 45 - 9
= 72 - 72
= 0.
Since, f(3) = 0, thus (x - 3) is a factor of f(x).
Now, dividing f(x) by x - 3,
$$\begin{array}{l} \phantom{x - 3)}{x^2 - 4x + 3} \\ x - 3\overline{\smash{\big)}x^3 - 7x^2 + 15x - 9} \\ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}3x^2} \\ \phantom{{x - 2}x^3-}-4x^2 + 15x \\ \phantom{{x - 2}x^3ok\space}\underline{\underset{+}{-}4x^2\underset{-}{+} 12x} \\ \phantom{{x - 2uo[ki]}x^3okklk\space}{3x - 9} \\ \phantom{{x - 2}x^3o;lmkb\k\space}\underline{\underset{-}{+}3x\underset{+}{-} 9} \\ \phantom{{x - 2}{x^,jo\3-2x^2\space}{-9x}}\times \end{array}$$
∴ x3 - 7x2 + 15x - 9 = (x - 3)(x2 - 4x + 3)
= (x - 3)(x2 - 3x - x + 3)
= (x - 3)[x(x - 3) - 1(x - 3)]
= (x - 3)(x - 1)(x - 3)
= (x - 3)2(x - 1).
Hence, x3 - 7x2 + 15x - 9 = (x - 3)2(x - 1).
Using factor theorem, show that (x - 4) is a factor of (2x3 + x2 - 26x - 40) and hence factorize (2x3 + x2 - 26x - 40).
Let f(x) = 2x3 + x2 - 26x - 40
Substituting x = 4 in f(x), we get :
f(4) = 2(4)3 + (4)2 - 26(4) - 40
= 2(64) + 16 - 104 - 40
= 128 + 16 - 104 - 40
= 144 - 144
= 0.
Since f(4) = 0, (x − 4) is a factor of 2x3 + x2 - 26x - 40
Now, dividing f(x) by x - 4,
$$\begin{array}{l} \phantom{x - 3)}{2x^2 + 9x + 10} \\ x - 4\overline{\smash{\big)}2x^3 + x^2 - 26x - 40} \\ \phantom{x - 2}\underline{\underset{-}{+}2x^3 \underset{+}{-}8x^2} \\ \phantom{{x - 2}x^3-}9x^2 - 26x \\ \phantom{{x -l2}x^3\space}\underline{\underset{-}{+}9x^2\underset{+}{-} 36x} \\ \phantom{{x - 2uo[ki]}x^3okk\space}{10x - 40} \\ \phantom{{x - 2}x^3o;lmkk\space}\underline{\underset{-}{+}10x\underset{+}{-} 40} \\ \phantom{{x - 2}{x^,j-k2x^2\space}{-9x}}\times \end{array}$$
∴ 2x3 + x2 - 26x - 40 = (x - 4)(2x2 + 9x + 10)
= (x - 4)(2x2 + 4x + 5x + 10)
= (x - 4)[2x(x + 2) + 5(x + 2)]
= (x - 4)(2x + 5)(x + 2)
Hence, 2x3 + x2 - 26x - 40 = (x - 4)(2x + 5)(x + 2).
Show that (x - 3) is a factor of (2x3 - 3x2 - 11x + 6) and hence factorize (2x3 - 3x2 - 11x + 6).
Let f(x) = 2x3 - 3x2 - 11x + 6
Substituting x = 3 in f(x), we get :
f(3) = 2(3)3 - 3(3)2 - 11(3) + 6
= 54 - 27 - 33 + 6
= 60 - 60
= 0.
Since f(3) = 0, thus (x − 3) is a factor of 2x3 - 3x2 - 11x + 6.
Now, dividing f(x) by x - 3,
$$\begin{array}{l} \phantom{x - 3)}{2x^2 + 3x - 2} \\ x - 3\overline{\smash{\big)}2x^3 - 3x^2 - 11x + 6} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}6x^2} \\ \phantom{{x - 2}x^,,,3-}3x^2 - 11x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}3x^2 \underset{+}{-} 9x} \\ \phantom{{x - 2euo[ki]}x^3okk\space}{-2x + 6} \\ \phantom{{x - 2}x^3o;llk]lmk,\space}\underline{\underset{+}{-}2x\underset{-}{+} 6} \\ \phantom{{x - 2}{x^,jo\ -k2x^2\space}{-9x}}\times \end{array}$$
∴ 2x3 - 3x2 - 11x + 6 = (x - 3)(2x2 + 3x - 2)
= (x - 3)(2x2 + 4x - x - 2)
= (x - 3)[2x(x + 2) - 1(x + 2)]
= (x - 3)(2x - 1)(x + 2).
Hence, 2x3 - 3x2 - 11x + 6 = (x - 3)(2x - 1)(x + 2).
Show that (3x + 2) is a factor of (6x3 + 13x2 - 4) and hence factorize (6x3 + 13x2 - 4).
Let, f(x) = 6x3 + 13x2 - 4.
Factor :
⇒ 3x + 2 = 0
⇒ 3x = -2
⇒ x = $-\frac{2}{3}$.
Substituting, x = $\Big(\frac{-2}{3}\Big)$ in f(x), we get :
$$\begin{aligned}f\Big(-\frac{2}{3}\Big) = 6\Big(-\frac{2}{3}\Big)^3 + 13\Big(-\frac{2}{3}\Big)^2 - 4 \\[1em] = 6\Big(-\frac{8}{27}\Big) + 13\Big(\frac{4}{9}\Big) - 4 \\[1em] = \Big(-\frac{48}{27}\Big) + \Big(\frac{52}{9}\Big) - 4 \\[1em] = \frac{-48 + 156 - 108}{27} \\[1em] = \frac{156 - 156}{27} \\[1em] = 0.\end{aligned}$$
Since, $f\Big(-\frac{2}{3}\Big)$ = 0, thus (3x + 2) is a factor of 6x3 + 13x2 - 4.
Now, dividing f(x) by (3x + 2),
$$\begin{array}{l} \phantom{x -;]/3)}{2x^2 + 3x - 2} \\ 3x + 2\overline{\smash{\big)}6x^3 + 13x^2 - 4} \\ \phantom{x - 2l}\underline{\underset{-}{}6x^3 \underset{-}{+}4x^2} \\ \phantom{{x - 2}ex^,,,3-}9x^2 \\ \phantom{{x e[[-l2}fx^3]\space}\underline{\underset{-}{+}9x^2 \underset{-}{+} 6x} \\ \phantom{{x - 2]euo[ki]}x^3ok\space}{-6x - 4} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{+}{-}6x\underset{+}{-} 4} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$$
6x3 + 13x2 - 4 = (3x + 2)(2x2 + 3x - 2)
= (3x + 2)(2x2 + 4x - x - 2)
= (3x + 2)[2x(x + 2) - 1(x + 2)]
= (3x + 2)(2x - 1)(x + 2)
Hence, 6x3 + 13x2 - 4 = (3x + 2)(2x - 1)(x + 2).
If 2x3 - 3x2 - 3x + 2 = (2x - 1)(x2 + ax + b)
(i) using Remainder and Factor theorem, find the value of ‘a’ and ‘b’.
(ii) hence, factorise the polynomial 2x3 - 3x2 - 3x + 2 completely.
(i) Given,
2x3 - 3x2 - 3x + 2 = (2x - 1)(x2 + ax + b)
Therefore, (2x - 1) is factor of 2x3 - 3x2 - 3x + 2.
Factorizing,
$$\begin{array}{l} \phantom{x - 3x)}{\quad x^2 - x - 2} \\ 2x - 1\overline{\smash{\big)}\quad 2x^3 - 3x^2 - 3x + 2 } \\ \phantom{x^2 + 4}\phantom(\underline{\underset{-}{+}2x^3 \underset{+}{-}x^2} \\ \phantom{x^2 + 3x - 7)} - 2x^2 - 3x \\ \phantom{x^2 + 3x - 5))}\underline{\underset{+}{-}2x^2 \underset{-}{+}x} \\ \phantom{x^2 + 3x - 5) + 24)}-4x + 2 \\ \phantom{{x^2 + 3x - 5) + 24 +}}\underline{\underset{+}{-}4x \underset{-}{+}2} \\ \phantom{{x^2 + 3x - 54)} + 2x + 7} \times\ \end{array}$$
∴ 2x3 - 3x2 - 3x + 2 = (2x - 1)(x2 - x - 2)
Comparing, x2 + ax + b with x2 - x - 2, we get:
a = -1 and b = -2.
Hence, a = -1 and b = -2.
(ii) From part (i),
⇒ 2x3 - 3x2 - 3x + 2 = (2x - 1)(x2 - x - 2)
= (2x - 1)(x2 - 2x + x - 2)
= (2x - 1)[x(x - 2) + 1(x - 2)]
= (2x - 1)(x + 1)(x - 2).
Hence, 2x3 - 3x2 - 3x + 2 = (2x - 1)(x + 1)(x - 2).
While factorizing a given polynomial, using remainder and factor theorem, a student finds that (2x + 1) is a factor of 2x3 + 7x2 + 2x – 3.
(i) Is the student’s solution correct stating that (2x + 1) is a factor of the given polynomial ?
(ii) Give a valid reason for your answer.
Also factorize the given polynomial completely.
Given,
⇒ 2x + 1 = 0
⇒ 2x = -1
⇒ x = $-\frac{1}{2}$
Substituting x = $-\frac{1}{2}$ in 2x3 + 7x2 + 2x – 3, we get:
$$\begin{aligned}\Rightarrow 2 \times \Big(-\frac{1}{2}\Big)^3 + 7 \times \Big(-\frac{1}{2}\Big)^2 + 2 \times \Big(-\frac{1}{2}\Big) - 3 \\[1em] \Rightarrow 2 \times -\frac{1}{8} + 7 \times \frac{1}{4} + (-1) - 3 \\[1em] \Rightarrow -\frac{1}{4} + \frac{7}{4} - 4 \\[1em] \Rightarrow \frac{-1 + 7}{4} - 4 \\[1em] \Rightarrow \frac{6}{4} - 4 \\[1em] \Rightarrow \frac{6 - 16}{4} \\[1em] \Rightarrow \frac{-10}{4} \\[1em] \Rightarrow \frac{-5}{2}.\end{aligned}$$
Since, remainder is not equal to zero.
Hence, (2x + 1) is not a factor of the given polynomial.
Substituting x = $\frac{1}{2}$ in 2x3 + 7x2 + 2x – 3, we get:
$$\begin{aligned}\Rightarrow 2 \times \Big(\frac{1}{2}\Big)^3 + 7 \times \Big(\frac{1}{2}\Big)^2 + 2 \times \Big(\frac{1}{2}\Big) - 3 \\[1em] \Rightarrow 2 \times \frac{1}{8} + 7 \times \frac{1}{4} + 1 - 3 \\[1em] \Rightarrow \frac{1}{4} + \frac{7}{4} - 2 \\[1em] \Rightarrow \frac{1 + 7}{4} - 2 \\[1em] \Rightarrow \frac{8}{4} - 2 \\[1em] \Rightarrow 2 - 2 \\[1em] \Rightarrow 0.\end{aligned}$$
Since, remainder is equal to zero. Thus, x - $\frac{1}{2}$ is the factor of the polynomial.
$$\begin{aligned}\Rightarrow x - \frac{1}{2} = 0 \\[1em] \Rightarrow x = \frac{1}{2} \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow 2x - 1.\end{aligned}$$
2x - 1 is factor of polynomial.
Dividing 2x3 + 7x2 + 2x – 3 by 2x - 1, we get:
$$\begin{array}{l} \phantom{x - 3x)}{\quad x^2 + 4x + 3} \\ 2x - 1\overline{\smash{\big)}\quad 2x^3 + 7x^2 + 2x – 3} \\ \phantom{x^2 +o 4}\phantom(\underline{\underset{-}{+}2x^3 \underset{+}{-}x^2} \\ \phantom{x^2 + 3x - 7)[][]} 8x^2 + 2x \\ \phantom{x^2 + 3x - 5))}\underline{\underset{-}{+}8x^2 \underset{+}{-}4x} \\ \phantom{x^2 + 3x - 5) + (24)}6x - 3 \\ \phantom{{x^2 + 3x - 5) + 24}}\underline{\underset{-}{+}6x \underset{+}{-}3} \\ \phantom{{x^2 + 3x - 54)} + 2x ,k}\times\ \end{array}$$
2x3 + 7x2 + 2x – 3 by 2x - 1 = (2x - 1)(x2 + 4x + 3)
= (2x - 1)[x2 + 3x + x + 3]
= (2x - 1)[x(x + 3) + 1(x + 3)]
= (2x - 1)(x + 1)(x + 3).
Hence, 2x3 + 7x2 + 2x – 3 = (2x - 1)(x + 1)(x + 3).
Using the factor theorem, show that (x - 2) is a factor of x3 + x2 - 4x - 4. Hence factorize the polynomial completely.
Let, f(x) = x3 + x2 - 4x - 4.
Factor :
⇒ x - 2 = 0
⇒ x = 2.
Substituting x = 2 in f(x), we get :
⇒ f(2) = (2)3 + (2)2 - 4(2) - 4
= 8 + 4 - 8 - 4
= 12 - 12
= 0.
Since, f(2) = 0, thus (x - 2) is a factor of (x3 + x2 - 4x - 4).
Now, dividing f(x) by (x - 2), we get :
$$\begin{array}{l} \phantom{x -13}{x^2 + 3x + 2} \\ x - 2\overline{\smash{\big)}x^3 + x^2 - 4x - 4} \\ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}2x^2} \\ \phantom{{x - 2}x^,.3-}3x^2 - 4x \\ \phantom{{x -l2}fx^l.\space}\underline{\underset{-}{+}3x^2 \underset{+}{-}6x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{2x - 4} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}2x\underset{+}{-}4} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$$
∴ x3 + x2 - 4x - 4 = (x - 2)(x2 + 3x + 2)
= (x - 2)(x2 + x + 2x + 2)
= (x - 2)[x(x + 1) + 2(x + 1)]
= (x - 2)(x + 2)(x + 1).
Hence, x3 + x2 - 4x - 4 = (x - 2)(x + 2)(x + 1).
If (x - 2) is a factor of 2x3 - x2 - px - 2,
(i) find the value of p
(ii) with the value of p, factorize the above expression completely.
(i) Let f(x) = 2x3 - x2 - px - 2
Since, (x − 2) is the factor, f(2) = 0.
⇒ 2(2)3 - (2)2 - p(2) - 2 = 0
⇒ 16 - 4 - 2p - 2 = 0
⇒ 10 - 2p = 0
⇒ 2p = 10
⇒ p = $\frac{10}{2}$
⇒ p = 5.
Hence, the value of p = 5.
(ii) f(x) = 2x3 - x2 - 5x - 2
Now, dividing f(x) by (x - 2), we get :
$$\begin{array}{l} \phantom{x -1 ]3)}{2x^2 + 3x + 1} \\ x - 2\overline{\smash{\big)}2x^3 - x^2 - 5x - 2} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}4x^2} \\ \phantom{{x - 2}x^,.3-}3x^2 - 5x \\ \phantom{{x -l2}fx^l.\space}\underline{\underset{-}{+}3x^2 \underset{+}{-}6x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{x - 2} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}x\underset{+}{-}2} \\ \phantom{{x - 2}{x^,jo-k2\space}{-9x}}\times \end{array}$$
2x3 - x2 - 5x - 2 = (x - 2)(2x2 + 3x + 1)
= (x - 2)(2x2 + 2x + x + 1)
= (x - 2)[2x(x + 1) + 1(x + 1)]
= (x - 2)(2x + 1)(x + 1)
Hence, 2x3 - x2 - 5x - 2 = (x - 2)(2x + 1)(x + 1).
Find the value of a, if (x - a) is a factor of the polynomial 3x3 + x2 - ax - 81.
Let, f(x) = 3x3 + x2 - ax - 81.
Factor :
⇒ x - a = 0
⇒ x = a.
Since (x − a) is the factor, thus f(a) = 0.
⇒ 3(a)3 + a2 - a(a) - 81 = 0
⇒ 3a3 + a2 - a2 - 81 = 0
⇒ 3a3 - 81 = 0
⇒ 3a3 = 81
⇒ a3 = $\frac{81}{3}$
⇒ a3 = 27
⇒ a = $\sqrt[3]{27}$
⇒ a = 3.
Hence, the value of a = 3.
Find the values of a and b, if (x - 1) and (x + 2) are both factors of (x3 + ax2 + bx - 6).
Let f(x) = x3 + ax2 + bx - 6
Since (x − 1) and (x + 2) are factors, by the factor theorem, f(1) = 0 and f(−2) = 0.
⇒ f(1) = 0
⇒ (1)3 + a(1)2 + b(1) - 6 = 0
⇒ 1 + a + b - 6 = 0
⇒ a + b - 5 = 0
⇒ a + b = 5 ....(1)
⇒ f(-2) = 0
⇒ (-2)3 + a(-2)2 + b(-2) - 6 = 0
⇒ -8 + 4a - 2b - 6 = 0
⇒ 4a - 2b = 14
⇒ 2(2a - b) = 14
⇒ 2a - b = $\frac{14}{2}$
⇒ 2a - b = 7 ....(2)
Adding equations (1) and (2), we get:
⇒ a + b + 2a - b = 5 + 7
⇒ 3a = 12
⇒ a = $\frac{12}{3}$
⇒ a = 4.
Substituting value of a in equation (1), we get :
⇒ 4 + b = 5
⇒ b = 5 - 4
⇒ b = 1.
Hence, the value of a = 4 and b = 1.
If (x + 2) and (x + 3) are factors of x3 + ax + b, find the values of a and b.
Let f(x) = x3 + ax + b
Since (x + 2) and (x + 3) are factors, by the factor theorem, f(−2) = 0 and f(−3) = 0.
⇒ f(-2) = 0
⇒ (-2)3 + a(-2) + b = 0
⇒ -8 - 2a + b = 0
⇒ -2a + b = 8 ....(1)
⇒ f(-3) = 0
⇒ (-3)3 + a(-3) + b = 0
⇒ -27 - 3a + b = 0
⇒ -3a + b = 27 ....(2)
Subtract equation (2) from equation (1), we get:
⇒ -2a + b - (-3a + b) = 8 - 27
⇒ -2a + 3a = -19
⇒ a = -19
Substituting value of a in equation (1), we get :
⇒ -2(-19) + b = 8
⇒ 38 + b = 8
⇒ b = 8 - 38
⇒ b = -30
Hence, the value of a = -19 and b = -30.
If (x3 + ax2 + bx + 6) has (x - 2) as a factor and leaves a remainder 3 when divided by (x - 3), find the values of a and b.
Let f(x) = x3 + ax2 + bx + 6
By factor theorem,
If, (x - 2) is a factor of f(x), then f(2) = 0.
⇒ (2)3 + a(2)2 + b(2) + 6 = 0
⇒ 8 + 4a + 2b + 6 = 0
⇒ 4a + 2b + 14 = 0
⇒ 2(2a + b + 7) = 0
⇒ 2a + b + 7 = 0
⇒ 2a + b = -7 ....(1)
Given,
On dividing f(x) by (x − 3), the remainder is 3.
By remainder theorem,
∴ f(3) = 3
⇒ (3)3 + a(3)2 + b(3) + 6 = 3
⇒ 27 + 9a + 3b + 6 = 3
⇒ 9a + 3b + 33 = 3
⇒ 9a + 3b = 3 - 33
⇒ 9a + 3b = -30
⇒ 3(3a + b) = -30
⇒ 3a + b = $-\frac{30}{3}$
⇒ 3a + b = -10 ....(2)
Subtracting equation (1) from equation (2),
⇒ 3a + b - (2a + b) = -10 -(-7)
⇒ 3a + b - 2a - b = -10 + 7
⇒ a = -3.
Substituting value of a in equation (1), we get :
⇒ 2(-3) + b = -7
⇒ -6 + b = -7
⇒ b = -7 + 6
⇒ b = -1.
Hence, the value of a = -3 and b = -1.
Using factor theorem, factorize the following:
x3 + 7x2 + 7x - 15
Let, f(x) = x3 + 7x2 + 7x - 15.
Substituting, x = 1 in f(x), we get :
f(1) = (1)3 + 7(1)2 + 7(1) - 15
= 1 + 7 + 7 - 15
= 0.
Since, f(1) = 0, thus (x - 1) is a factor of f(x).
Dividing, f(x) by (x - 1), we get :
$$\begin{array}{l} \phantom{x - ]3)}{x^2 + 8x + 15} \\ x - 1\overline{\smash{\big)}x^3 + 7x^2 + 7x - 15} \\ \phantom{x - 2l}\underline{\underset{-}{}x^3 \underset{+}{-}x^2} \\ \phantom{{x - 2}x^,,,3-}8x^2 + 7x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}8x^2 \underset{+}{-} 8x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{15x - 15} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}15x\underset{+}{-} 15} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$$
∴ x3 + 7x2 + 7x - 15 = (x - 1)(x2 + 8x + 15)
= (x - 1)(x2 + 3x + 5x + 15)
= (x - 1)[x(x + 3) + 5(x + 3)]
= (x - 1)(x + 5)(x + 3).
Hence, x3 + 7x2 + 7x - 15 = (x - 1)(x + 5)(x + 3).
Using factor theorem, factorize the following:
6x3 - 7x2 - 11x + 12
Let, f(x) = 6x3 - 7x2 - 11x + 12.
Substituting, x = 1 in f(x), we get :
f(1) = 6(1)3 - 7(1)2 - 11(1) + 12
= 6 - 7 - 11 + 12
= 0
Since, f(1) = 0, (x - 1) is a factor of f(x).
Dividing f(x) by (x - 1), we get :
$$\begin{array}{l} \phantom{x - ]3)}{6x^2 - x - 12} \\ x - 1\overline{\smash{\big)}6x^3 - 7x^2 - 11x + 12} \\ \phantom{x - 2l}\underline{\underset{-}{}6x^3 \underset{+}{-}6x^2} \\ \phantom{{x - 2}x^,,,3-}-x^2 - 11x \\ \phantom{{x -k.l2}fx^3]\space}\underline{\underset{+}{-}x^2 \underset{-}{+}x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{-12x + 12} \\ \phantom{{x - 2}x^3o;llk]lmk,\space}\underline{\underset{+}{-}12x\underset{-}{+}12} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$$
∴ 6x3 − 7x2 − 11x + 12 = (x − 1)(6x2 − x − 12)
= (x − 1)(6x2 − 9x + 8x − 12)
= (x − 1)[3x(2x − 3) + 4(2x − 3)]
= (x − 1)(2x − 3)(3x + 4)
Hence, 6x3 − 7x2 − 11x + 12 = (x − 1)(2x − 3)(3x + 4).
Using factor theorem, factorize the following:
2x3 + 3x2 − 9x − 10
Let, f(x) = 2x3 + 3x2 − 9x − 10.
Substituting, x = 2 in f(x), we get :
f(2) = 2(2)3 + 3(2)2 − 9(2) − 10
= 16 + 12 − 18 − 10
= 0.
Since, f(2) = 0, (x − 2) is a factor of f(x).
Dividing f(x) by (x − 2), we get :
$$\begin{array}{l} \phantom{x - ]k3)}{2x^2 + 7x + 5} \\ x - 2\overline{\smash{\big)}2x^3 + 3x^2 - 9x - 10} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}4x^2} \\ \phantom{{x - 2}x^,,,3-}7x^2 - 9x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}7x^2 \underset{+}{-}14x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{5x - 10} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}5x\underset{+}{-}10} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$$
∴ 2x3 + 3x2 − 9x − 10 = (x − 2)(2x2 + 7x + 5)
= (x − 2)(2x2 + 5x + 2x + 5)
= (x − 2)[x(2x + 5) + 1(2x + 5)]
= (x − 2)(2x + 5)(x + 1)
Hence, 2x3 + 3x2 − 9x − 10 = (x − 2)(2x + 5)(x + 1).
Using factor theorem, factorize the following:
2x3 + 19x2 + 38x + 21
Let, f(x) = 2x3 + 19x2 + 38x + 21.
Substituting, x = −1 in f(x), we get :
f(-1) = 2(-1)3 + 19(-1)2 + 38(-1) + 21
= 2(-1) + 19(1) - 38 + 21
= -2 + 19 - 38 + 21
= 0.
Since, f(−1) = 0, (x + 1) is a factor of f(x).
Dividing f(x) by (x + 1), we get :
$$\begin{array}{l} \phantom{x - ]3)}{2x^2 + 17x + 21} \\ x + 1\overline{\smash{\big)}2x^3 + 19x^2 + 38x + 21} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{-}{+}2x^2} \\ \phantom{{x - 2}x^,,,3-}17x^2 + 38x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}17x^2 \underset{-}{+}17x} \\ \phantom{{x - ll2]euo[ki]}x^3okk\space}{21x + 21} \\ \phantom{{x - 2}x^3o;lklk]lmk\space}\underline{\underset{-}{+}21x\underset{-}{+} 21} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$$
∴ 2x3 + 19x2 + 38x + 21 = (x + 1)(2x2 + 17x + 21)
= (x + 1)(2x2 + 14x + 3x + 21)
= (x + 1)[2x(x + 7) + 3(x + 7)]
= (x + 1)(x + 7)(2x + 3)
Hence, 2x3 + 19x2 + 38x + 21 = (x + 1)(x + 7)(2x + 3).
Using factor theorem, factorize the following:
3x3 + 2x2 - 19x + 6
Let, f(x) = 3x3 + 2x2 - 19x + 6.
Substituting, x = 2 in f(x), we get :
f(2) = 3(2)3 + 2(2)2 - 19(2) + 6
= 3(8) + 2(4) - 38 + 6
= 24 + 8 - 38 + 6
= 0.
Since, f(2) = 0, thus (x - 2) is a factor of f(x).
Dividing f(x) by (x - 2), we get :
$$\begin{array}{l} \phantom{x -.]3)}{3x^2 + 8x - 3} \\ x - 2\overline{\smash{\big)}3x^3 + 2x^2 - 19x + 6} \\ \phantom{x - l}\underline{\underset{-}{}3x^3 \underset{+}{-}6x^2} \\ \phantom{{x - 2}x^,,,3-}8x^2 - 19x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}8x^2 \underset{+}{-}16x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{-3x + 6} \\ \phantom{{x - 2}x,'^3o;llk]lmk\space}\underline{\underset{+}{-}3x\underset{-}{+}6} \\ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$$
∴ 3x3 + 2x2 - 19x + 6 = (x - 2)(3x2 + 8x - 3)
= (x - 2)(3x2 + 9x - x - 3)
= (x - 2)[3x(x + 3) - 1(x + 3)]
= (x - 2)(3x - 1)(x + 3).
Hence, 3x3 + 2x2 − 19x + 6 = (x − 2)(3x − 1)(x + 3).
Using factor theorem, factorize the following:
2x3 + x2 - 13x + 6
Let, f(x) = 2x3 + x2 - 13x + 6.
Substituting, x = 2 in f(x) we get :
f(2) = 2(2)3 + (2)2 - 13(2) + 6
= 2(8) + 4 - 26 + 6
= 16 + 4 - 26 + 6
= 0.
Since, f(2) = 0, thus (x - 2) is factor of f(x).
Dividing, f(x) by (x - 2), we get :
$$\begin{array}{l} \phantom{x - ]3)}{2x^2 + 5x - 3} \\ x - 2\overline{\smash{\big)}2x^3 + x^2 - 13x + 6} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}4x^2} \\ \phantom{{x - 2}x^,,,3-}5x^2 - 13x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}5x^2 \underset{+}{-}10x} \\ \phantom{{x - 2]euo[ki]}x^3o.\space}{-3x + 6} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{+}{-}3x\underset{-}{+}6} \\ \phantom{{x - 2}{x^,jo-k2x^2\space}{-9x}}\times \end{array}$$
∴ 2x3 + x2 - 13x + 6 = (x - 2)(2x2 + 5x - 3)
= (x - 2)(2x2 + 6x - x - 3)
= (x - 2)[2x(x + 3) - 1(x + 3)]
= (x - 2)(2x - 1)(x + 3).
Hence, 2x3 + x2 − 13x + 6 = (x − 2)(2x − 1)(x + 3).
Using factor theorem, factorize the following:
2x3 − x2 − 13x − 6
Let, f(x) = 2x3 − x2 − 13x − 6.
Substituting, x = 3 in f(x) we get :
f(3) = 2(3)3 − (3)2 − 13(3) − 6
= 2(27) − 9 − 39 − 6
= 54 − 9 − 39 − 6
= 0.
Since, f(3) = 0, thus (x − 3) is factor of f(x).
Dividing, f(x) by (x − 3), we get :
$$\begin{array}{l} \phantom{x - ]3)}{2x^2 + 5x + 2} \\ x - 3\overline{\smash{\big)}2x^3 - x^2 - 13x - 6} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}6x^2} \\ \phantom{{x - 2}x^,,,3-}5x^2 - 13x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}5x^2 \underset{+}{-}15x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{2x - 6} \\ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}2x\underset{+}{-}6} \\ \phantom{{x - 2}{x^,jo-k2x^2\space}{-9x}}\times \end{array}$$
∴ 2x3 − x2 − 13x − 6 = (x − 3)(2x2 + 5x + 2)
= (x − 3)(2x2 + 4x + x + 2)
= (x − 3)[2(x + 2) + 1(x + 2)]
= (x − 3)(2x + 1)(x + 2).
Hence, 2x3 − x2 − 13x − 6 = (x − 3)(2x + 1)(x + 2).
If (x - 2) is a factor of (x3 + 2x2 - kx + 10), find the value of k. Hence, determine whether (x + 5) is also a factor of the given expression.
Let, f(x) = x3 + 2x2 - kx + 10.
Since, x - 2 is factor of f(x), thus f(2) = 0.
∴ (2)3 + 2(2)2 - k(2) + 10 = 0
⇒ 8 + 2(4) - 2k + 10 = 0
⇒ 8 + 8 - 2k + 10 = 0
⇒ 26 - 2k = 0
⇒ 2k = 26
⇒ k = $\frac{26}{2}$
⇒ k = 13.
f(x) = x3 + 2x2 - 13x + 10
⇒ x + 5 = 0
⇒ x = -5.
f(-5) = (-5)3 + 2(-5)2 - 13(-5) + 10
= -125 + 2(25) + 65 + 10
= -125 + 50 + 65 + 10
= 125 - 125
= 0.
Since f(−5) = 0, thus (x + 5) is a factor of f(x).
Hence, value of k = 13 and x - 5 is factor of x3 + 2x2 - 13x + 10.
Using the remainder and factor theorems, factorize the polynomial.
x3 + 10x2 - 37x + 26
Let, f(x) = x3 + 10x2 - 37x + 26.
Substituting, x = 1 in f(x) we get :
f(1) = (1)3 + 10(1)2 - 37(1) + 26
= 1 + 10 - 37(1) + 26
= 37 - 37
= 0.
Since, f(1) = 0, thus (x − 1) is a factor of f(x).
Dividing, x3 + 10x2 - 37x + 26 by (x - 1), we get :
$$\begin{array}{l} \phantom{x - ]3)}{x^2 + 11x - 26} \\ x - 1\overline{\smash{\big)}x^3 + 10x^2 - 37x + 26} \\ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}x^2} \\ \phantom{{x - 2}x^,,,3-}11x^2 - 37x \\ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}11x^2 \underset{+}{-}11x} \\ \phantom{{x - 2]euo[ki]}x^3okk\space}{-26x + 26} \\ \phantom{{x - 2}x^3o;llk]lmttk\space}\underline{\underset{+}{-}26x\underset{-}{+}26} \\ \phantom{{x - 2}{x^,jo-k2x^mm2\space}{-9x}}\times \end{array}$$
∴ x3 + 10x2 - 37x + 26 = (x - 1)(x2 + 11x - 26)
= (x - 1)(x2 + 13x - 2x - 26)
= (x - 1)[x(x + 13) - 2(x + 13)]
= (x - 1)(x + 13)(x - 2).
Hence, x3 + 10x2 - 37x + 26 = (x - 1)(x + 13)(x - 2).
If (x - 2) is a factor of the expression 2x3 + ax2 + bx - 14 and when the expression is divided by (x - 3), it leaves a remainder 52, find the values of a and b.
Let f(x) = 2x3 + ax2 + bx - 14
Since, (x − 2) is a factor of f(x) then f(2) = 0.
⇒ 2(2)3 + a(2)2 + b(2) - 14 = 0
⇒ 2(8) + 4a + 2b - 14 = 0
⇒ 16 + 4a + 2b - 14 = 0
⇒ 4a + 2b + 2 = 0
⇒ 4a + 2b = -2
⇒ 2(2a + b) = -2
⇒ 2a + b = $\frac{-2}{2}$
⇒ 2a + b = -1 ....(1)
On dividing f(x) by (x − 3), the remainder is 52,
By remainder theorem,
⇒ f(3) = 52
⇒ 2(3)3 + a(3)2 + b(3) - 14 = 52
⇒ 2(27) + 9a + 3b - 14 = 52
⇒ 54 + 9a + 3b - 14 = 52
⇒ 9a + 3b + 40 = 52
⇒ 9a + 3b = 52 - 40
⇒ 9a + 3b = 12
⇒ 3(3a + b) = 12
⇒ 3a + b = $\frac{12}{3}$
⇒ 3a + b = 4 ....(2)
Subtracting equation (1) from (2), we get :
⇒ 3a + b - (2a + b) = 4 - (-1)
⇒ a = 4 + 1
⇒ a = 5.
Substituting a = 5 in equation (1), we get :
⇒ 2(5) + b = -1
⇒ 10 + b = -1
⇒ b = -1 - 10
⇒ b = -11.
Hence, the value of a = 5 and b = -11.
The polynomial 3x3 + 8x2 - 15x + k has (x - 1) as a factor. Find the value of k. Hence factorize the resulting polynomial completely.
⇒ x - 1 = 0
⇒ x = 1.
Given, (x - 1) is a factor of 3x3 + 8x2 - 15x + k.
Thus, on substituting x = 1 in 3x3 + 8x2 - 15x + k, the remainder will be zero.
⇒ 3.(1)3 + 8.(1)2 - 15(1) + k = 0
⇒ 3.1 + 8.1 - 15 + k = 0
⇒ 3 + 8 - 15 + k = 0
⇒ 11 - 15 + k = 0
⇒ k - 4 = 0
⇒ k = 4.
Polynomial = 3x3 + 8x2 - 15x + 4
On dividing (3x3 + 8x2 - 15x + 4) by (x - 1), we get :
$$\begin{array}{l} \phantom{x - 1)}{\quad 3x^2 + 11x - 4} \\ x - 1\overline{\smash{\big)}\quad 3x^3 + 8x^2 - 15x + 4} \\ \phantom{x - 1)}\phantom{)}\underline{\underset{-}{+}3x^3 \underset{+}{-}3x^2} \\ \phantom{{x - 1}31x^3-2}11x^2 - 15x \\ \phantom{{x - 1)}x^3-2}\underline{\underset{-}{+}11x^2 \underset{+}{-} 11x} \\ \phantom{{x - 1)}31x^3-2+1}-4x + 4 \\ \phantom{{x - 1)}31x^3-2+11}\underline{\underset{+}{-}4x \underset{-}{+} 4} \\ \phantom{{x - 1)}31x^3-2+11+1}\times \end{array}$$
⇒ 3x3 + 8x2 - 15x + 4 = (x - 1)(3x2 + 11x - 4)
= (x - 1)[3x2 + 12x - x - 4]
= (x - 1)[3x(x + 4) - 1(x + 4)]
= (x - 1)(3x - 1)(x + 4).
Hence, 3x3 + 8x2 - 15x + 4 = (x - 1)(3x - 1)(x + 4).
It is given that (x − 2) is a factor of polynomial 2x3 − 7x2 + kx − 2.
Find:
(i) the value of ‘k’.
(ii) hence, factorise the resulting polynomial completely.
(i) Since (x − 2) is a factor of 2x3 − 7x2 + kx − 2.
Thus, on substituting x = 2, in 2x3 − 7x2 + kx − 2, the remainder will be equal to zero.
⇒ 2(2)3 − 7(2)2 + k(2) − 2 = 0
⇒ 16 − 28 + 2k − 2 = 0
⇒ −14 + 2k = 0
⇒ 2k = 14
⇒ k = $\frac{14}{2}$
⇒ k = 7.
Hence, k = 7.
(ii) Substituting k = 7 in 2x3 − 7x2 + kx − 2, we get :
Polynomial : 2x3 − 7x2 + 7x − 2.
Dividing 2x3 − 7x2 + 7x − 2 by x - 2, we get :
$$\begin{array}{l} \phantom{x - 31}{\quad2x^2 - 3x + 1} \\ x - 2\overline{\smash{\big)}\quad 2x^3 − 7x^2 + 7x − 2} \\ \phantom{x^2 + 4}\phantom(\underline{\underset{-}{+}2x^3 \underset{+}{-}4x^2} \\ \phantom{x^2 + 3x - 7)} - 3x^2 + 7x \\ \phantom{x^2 + 3x - 5))}\underline{\underset{+}{-}3x^2 \underset{-}{+}6x} \\ \phantom{x^2 + 3x - 5) + 24x ())}x - 2 \\ \phantom{{x^2 + 3x - 5) + 24 +)}}\underline{\underset{-}{+}x \underset{+}{-}2} \\ \phantom{{x^2 + 3x - 54)} + 2x + 7} \times \end{array}$$
⇒ 2x3 − 7x2 + 7x − 2 = (x - 2)(2x2 - 3x + 1)
⇒ 2x3 − 7x2 + 7x − 2 = (x - 2)[2x2 - 2x - x + 1]
⇒ 2x3 − 7x2 + 7x − 2 = (x - 2)[2x(x - 1) - 1(x - 1)]
⇒ 2x3 − 7x2 + 7x − 2 = (x - 2)(2x - 1)(x - 1).
Hence, 2x3 − 7x2 + 7x − 2 = (x - 2)(2x - 1)(x - 1).
A polynomial in 'x' is divided by (x - a) and for (x - a) to be a factor of this polynomial, the remainder should be :
For (x - a) to be a factor of a polynomial, the remainder should be equal to zero.
Hence, Option 2 is the correct option.
If p(x) = x + 4, then p(x) + p(-x) = ?
Given,
⇒ p(x) = x + 4
⇒ p(-x) = -x + 4
⇒ p(x) + p(-x) = x + 4 - x + 4
= 4 + 4
= 8.
Hence, option 3 is the correct option.
If p(x) = x2 - $2\sqrt{2}x + 1, p(2\sqrt2)$ = ?
Given,
⇒ p(x) = x2 - $2\sqrt{2}x$ + 1
$$\begin{aligned}\Rightarrow p(2\sqrt2) = (2\sqrt2)^2 - (2\sqrt2)(2\sqrt2) + 1 \\[1em] = 2^2 \times 2 - 4 \times 2 + 1 \\[1em] = 8 - 8 + 1 \\[1em] = 1.\end{aligned}$$
Hence, option 2 is the correct option.
If f(x) = 3x - 5x2 - 1, then f(-1) = ?
Given,
⇒ f(x) = 3x - 5x2 - 1
⇒ f(-1) = 3(-1) - 5(-1)2 - 1
= -3 - 5 - 1
= -9.
Hence, option 4 is the correct option.
If (x101 + 101) is divided by (x + 1), then the remainder is:
The Remainder Theorem states that when a polynomial f(x) is divided by (x - a), the remainder is f(a).
Given,
f(x) = x101 + 101
f(-1) = (-1)101 + 101
= -1 + 101
= 100.
Hence, option 2 is the correct option.
If (3x3 - 5x2 + 3x - 7) is divided by (x - 2), then the remainder is :
By remainder theorem,
If a polynomial f(x) is divided by (x - a), remainder = f(a).
Let f(x) = 3x3 - 5x2 + 3x - 7
On dividing f(x) by x - 2, remainder = f(2).
f(2) = 3(2)3 - 5(2)2 + 3(2) - 7
= 3(8) - 5(4) + 6 - 7
= 24 - 20 + 6 - 7
= 3.
Hence, option 3 is the correct option.
If f(x) = x4 - ax3 + x2 - ax is divided by (x - a), then the remainder is:
By remainder theorem,
If a polynomial f(x) is divided by (x - a), remainder = f(a).
Given,
f(x) = x4 - ax3 + x2 - ax.
On dividing f(x) by x - a, remainder = f(a).
f(a) = a4 - a(a)3 + (a)2 - a(a)
= a4 - a4 + a2 - a2
= 0.
Hence, option 1 is the correct option.
When f(x) = x3 + ax2 + 2x + a is divided by (x + a), then the remainder is :
By remainder theorem,
If a polynomial f(x) is divided by (x - a), remainder = f(a).
Given,
f(x) = x3 + ax2 + 2x + a
On dividing f(x) by x + a, remainder = f(-a)
f(-a) = (-a)3 + a(-a)2 + 2(-a) + a
= -a3 + a3 - 2a + a
= -a.
Hence, option 2 is the correct option.
If (x2 - 7x + a) leaves a remainder 1 when divided by (x + 1), then the value of a is :
Given,
Let f(x) = x2 - 7x + a
Given,
On dividing f(x) by x + 1, remainder is 1.
By remainder theorem, remainder = f(-1).
⇒ f(-1) = 1
⇒ (-1)2 - 7(-1) + a = 1
⇒ 1 + 7 + a = 1
⇒ a = 1 - 8
⇒ a = -7.
Hence, option 3 is the correct option.
In the division of a cubic polynomial f(x) by a linear polynomial, the remainder is f(-2). Then the divisor must be:
By remainder theorem,
If a polynomial f(x) is divided by (x - a), remainder = f(a).
The remainder is given as f(-2).
⇒ x = -2
⇒ x + 2 = 0.
Hence, option 2 is the correct option.
If p(t) = t2 - t - 2, then the value of $p\Big(-\frac{1}{3}\Big)$ is :
Given,
p(t) = t2 - t - 2
$$\begin{aligned}\Rightarrow p\Big(-\frac{1}{3}\Big) = \Big(-\frac{1}{3}\Big)^2 - \Big(-\frac{1}{3}\Big) - 2 \\[1em] = \Big(\frac{1}{9}\Big) + \frac{1}{3} - 2 \\[1em] = \frac{1 + 1 \times 3 - 2 \times 9}{9} \\[1em] = \frac{1 + 3 - 18}{9} \\[1em] = -\frac{14}{9}.\end{aligned}$$
Hence, option 3 is the correct option.
If f(x) = 4x3 - 3x2 + 5 is divided by (2x + 1), then the remainder is :
Divisor :
⇒ 2x + 1 = 0
⇒ 2x = -1
⇒ x = $-\frac{1}{2}$.
Given,
f(x) = 4x3 - 3x2 + 5
By remainder theorem,
On dividing f(x) by 2x + 1, remainder = $f\Big(-\frac{1}{2}\Big)$.
$$\begin{aligned}\Rightarrow f\Big(-\frac{1}{2}\Big) = 4\Big(-\frac{1}{2}\Big)^3 - 3\Big(-\frac{1}{2}\Big)^2 + 5 \\[1em] = 4\Big(-\frac{1}{8}\Big) - 3\Big(\frac{1}{4}\Big) + 5 \\[1em] = -\frac{1}{2} - \frac{3}{4} + 5 \\[1em] = \frac{-1 \times 2 - 3 + 5 \times 4}{4} \\[1em] = \frac{-2 - 3 + 20}{4} \\[1em] = \frac{-5 + 20}{4} \\[1em] = \frac{15}{4}.\end{aligned}$$
Hence, option 4 is the correct option.
If (x50 - 1) is divided by (x - 1), then the remainder is :
Let, f(x) = x50 - 1.
By remainder theorem,
On dividing f(x) by x - 1, remainder = f(1).
⇒ f(1) = (1)50 - 1
= 1 - 1
= 0.
Hence, option 1 is the correct option.
When 2x2 - 3kx + k is divided by (x + 2), then the remainder obtained is 5. The value of k is:
Let, f(x) = 2x2 - 3kx + k
By remainder theorem,
On dividing f(x) by x + 2, remainder = f(-2).
Given,
Remainder = 5.
⇒ f(-2) = 5
⇒ 2(-2)2 - 3k(-2) + k = 5
⇒ 2(4) + 6k + k = 5
⇒ 8 + 7k = 5
⇒ 7k = 5 - 8
⇒ 7k = -3
⇒ k = $-\frac{3}{7}$.
Hence, option 3 is the correct option.
If the polynomial p(x) = x3 - 4kx + 3 is divided by (2x + 1), then the remainder obtained is -3. The value of k is:
Let, f(x) = x3 - 4kx + 3
⇒ 2x + 1 = 0
⇒ 2x = -1
⇒ x = $-\frac{1}{2}$
By remainder theorem,
On dividing f(x) by 2x + 1, remainder = $f\Big(-\frac{1}{2}\Big)$.
Given,
Remainder = -3
$$\begin{aligned}\Rightarrow f\Big(-\frac{1}{2}\Big) = -3 \\[1em] \Rightarrow \Big(-\frac{1}{2}\Big)^3 - 4k\Big(-\frac{1}{2}\Big) + 3 = -3 \\[1em] \Rightarrow \Big(-\frac{1}{8}\Big) + \frac{4k}{2} = - 3 - 3 \\[1em] \Rightarrow -\frac{1}{8} + 2k = - 6 \\[1em] \Rightarrow 2k = -6 + \frac{1}{8}\\[1em] \Rightarrow 2k = \frac{-6 \times 8 + 1}{8} \\[1em] \Rightarrow k = \frac{-48 + 1}{8 \times 2} \\[1em] \Rightarrow k = -\frac{47}{16}.\end{aligned}$$
Hence, option 4 is the correct option.
If (x + 5) is a factor of f(x) = x3 - 20x + 5k, then k = ?
Let, f(x) = x3 - 20x + 5k.
By factor theorem,
If (x + 5) is a factor of f(x), then f(-5) = 0.
⇒ (-5)3 - 20(-5) + 5k = 0
⇒ -125 + 100 + 5k = 0
⇒ -25 + 5k = 0
⇒ 5k = 25
⇒ k = $\frac{25}{5}$
⇒ k = 5.
Hence, option 3 is the correct option.
For what value of k is the polynomial f(x) = 2x3 - kx2 + 3x + 10 exactly divisible by (x + 2)?
Let, f(x) = 2x3 - kx2 + 3x + 10
By the Factor Theorem,
If (x + 2) is a factor of f(x), then f(-2) = 0.
⇒ 2(-2)3 - k(-2)2 + 3(-2) + 10 = 0
⇒ 2(-8) - k(4) - 6 + 10 = 0
⇒ -16 - 4k + 4 = 0
⇒ -4k - 12 = 0
⇒ 4k = -12
⇒ k = $-\frac{12}{4}$
⇒ k = -3.
Hence, option 2 is the correct option.
If (x50 + 2x49 + k) is divisible by (x + 1), then the value of k is:
Let, f(x) = x50 + 2x49 + k
By the Factor Theorem,
If (x + 1) is a factor of f(x), then f(-1) = 0.
⇒ (-1)50 + 2(-1)49 + k = 0
⇒ 1 + 2(-1) + k = 0
⇒ 1 - 2 + k = 0
⇒ k - 1 = 0
⇒ k = 1.
Hence, option 3 is the correct option.
The value of p for which (x - p) is a factor of x3 - px2 + x + 5 is:
Given,
f(x) = x3 - px2 + x + 5
⇒ x - p = 0
⇒ x = p.
Since, (x - p) is the factor of f(x), thus f(p) = 0.
f(p) = (p)3 - p(p)2 + p + 5
⇒ 0 = p3 - p3 + p + 5
⇒ 0 = p + 5
⇒ p = -5.
Hence, option 1 is the correct option.
If x - 2 is a factor of x3 - kx - 12, then the value of k is :
By factor theorem,
If x - a is a factor of f(x), then f(a) = 0.
Given,
x - 2 is a factor of x3 - kx - 12
⇒ x - 2 = 0
⇒ x = 2.
∴ 23 - 2k - 12 = 0
⇒ 8 - 2k - 12 = 0
⇒ -2k - 4 = 0
⇒ 2k = -4
⇒ k = $-\frac{4}{2}$ = -2.
Hence, Option 3 is the correct option.
For a polynomial f(x), f(-1) and f(2) are both equal to zero. Which of the following is a factor of f(x)?
By the Factor Theorem,
If f(-1) = 0, (x + 1) is a factor of f(x).
If f(2) = 0, (x - 2) is a factor of f(x).
Since both are the factors of f(x). Multiplying both the factors,
⇒ (x + 1)(x - 2)
⇒ x2 - 2x + x - 2
⇒ x2 - x - 2.
Hence, option 2 is the correct option.
(x + 1) is a factor of the polynomial :
Let, f(x) = x3 + 2x2 - x - 2
f(-1) = (-1)3 + 2(-1)2 - (-1) - 2
= (-1) + 2(1) + 1 - 2
= -1 + 2 + 1 - 2
= 3 - 3
= 0.
Since f(-1) = 0,
Thus, (x + 1) is factor of x3 + 2x2 - x - 2.
Hence, option 2 is the correct option.
A polynomial in x is x3 + 5x2 - kx - 24. Which of the following is a factor of the given polynomial so that the value of k is 2?
By factor theorem,
(x - a) is a factor of f(x), if f(a) = 0.
Given,
Polynomial = x3 + 5x2 - kx - 24
If k = 2, then :
Polynomial = x3 + 5x2 - 2x - 24.
Dividing polynomial by (x + 4) or substituting -4 in polynomial, we get :
⇒ (-4)3 + 5(-4)2 - 2(-4) - 24
⇒ -64 + 5(16) + 8 - 24
⇒ -64 + 80 + 8 - 24
⇒ 88 - 88
⇒ 0.
Since, on substituting -4 in polynomial, we get remainder = 0.
∴ (x + 4) is the factor of x3 + 5x2 - kx - 24, when k = 2.
Hence, Option 3 is the correct option.
What number should be subtracted from 2x3 - 5x2 + 5x, so that the resulting polynomial has (2x - 3) as a factor?
Let number to be subtracted be a, then resulting polynomial:
f(x) = 2x3 - 5x2 + 5x - a,
⇒ 2x - 3 = 0
⇒ 2x = 3
⇒ x = $\frac{3}{2}$.
If (2x - 3) is a factor of f(x), then $f\Big(\frac{3}{2}\Big) = 0$.
$$\begin{aligned}\Rightarrow 2\Big(\frac{3}{2}\Big)^3 - 5\Big(\frac{3}{2}\Big)^2 + 5\Big(\frac{3}{2}\Big) - a = 0\\[1em] \Rightarrow 2\Big(\frac{27}{8}\Big) - 5\Big(\frac{9}{4}\Big) + 5\Big(\frac{3}{2}\Big) - a = 0\\[1em] \Rightarrow \frac{2 \times 27}{8} - \frac{5 \times 9}{4} + \frac{3 \times 5}{2} -a = 0\\[1em] \Rightarrow \frac{27}{4} - \frac{45}{4} + \frac{15}{2} - a = 0\\[1em] \Rightarrow \frac{27 - 45 + 15 \times 2}{4} -a = 0\\[1em] \Rightarrow \frac{- 18 + 30}{4} -a = 0 \\[1em] \Rightarrow \frac{12}{4} - a = 0 \\[1em] \Rightarrow 3 -a =0\\[1em] \Rightarrow a = 3.\end{aligned}$$
If 3 is subtracted from 2x3 - 5x2 + 5x, then (2x - 3) is a factor.
Hence, option 4 is the correct option.
If (x2 + ax + b) is divided by (x + c), then the remainder is :
Given,
f(x) = x2 + ax + b
By remainder theorem,
On dividing f(x) by (x + c), remainder = f(-c).
⇒ f(-c) = (-c)2 + a(-c) + b
= c2 - ac + b.
Hence, option 4 is the correct option.
(x - 2) and (x + 2) are the factors of x3 + x2 - 4x - 4. The third factor of the given polynomial is :
Given,
(x - 2) and (x + 2) are the factors of x3 + x2 - 4x - 4.
⇒ (x - 2)(x + 2) = x2 - 4
Thus, x2 - 4 is also the factor of x3 + x2 - 4x - 4.
$$\begin{array}{l} \phantom{x - 3x)}{\quad x + 1} \\ x^2 - 4\overline{\smash{\big)}\quad x^3 + x^2 - 4x - 4 } \\ \phantom{x^2 + n}\phantom(\underline{\underset{-}{+}x^3 \underset{+}{-}4x} \\ \phantom{x^2 + 3x -])7 + o)()} x^2 - 4 \\ \phantom{x^2 + 3x - ((5) + )}\underline{\underset{-}{+}x^2 \underset{+}{-}4} \\ \phantom{x^2 + 3x - 5) + 24x )}\times \end{array}$$
Thus, x3 + x2 - 4x - 4 = (x2 - 4)(x + 1)
= (x + 2)(x - 2)(x + 1).
Hence, option 3 is the correct option.
For two polynomials f(x) and g(x), (x - a) and (x - b) are their respective factors. Which of the following is true?
By the Factor Theorem,
If (x - a) is a factor of f(x), then f(a) = 0.
If (x - b) is a factor of g(x), then g(b) = 0.
Thus, f(a) + g(b) = 0.
Hence, option 3 is the correct option.
If the polynomial 2x3 + 3x2 - 2x - 3 is completely divisible by (2x + a), and the quotient is equal to (x2 - 1), then one of the values of a is :
Given,
The polynomial 2x3 + 3x2 - 2x - 3 is completely divisible by (2x + a) and quotient is equal to (x2 - 1).
∴ 2x3 + 3x2 - 2x - 3 = (2x + a)(x2 - 1)
⇒ 2x3 + 3x2 - 2x - 3 = 2x3 - 2x + ax2 - a
⇒ 2x3 - 2x3 + 3x2 - 2x + 2x - 3 = ax2 - a
⇒ 3x2 - 3 = ax2 - a
From above equation,
a = 3.
Hence, Option 4 is the correct option.
Assertion (A): When a polynomial f(x) is divided by (3x + 4), then the remainder is $f\Big(-\frac{3}{4}\Big)$.
Reason (R): Remainder theorem states that when a polynomial f(x) is divided by (x - α), then the remainder is f(α).
Remainder Theorem states that when a polynomial f(x) is divided by a linear factor (x − α), the remainder is f(α).
∴ Reason (R) is true.
Divisor :
⇒ 3x + 4 = 0
⇒ 3x = -4
⇒ x = $-\frac{4}{3}$
Thus, when a polynomial f(x) is divided by (3x + 4), then the remainder is $f\Big(-\frac{4}{3}\Big)$.
∴ Assertion (A) is false.
Hence, option 4 is the correct option.
Assertion (A): (x - 1) is a factor of x3 + 2x2 - x - 2.
Reason (R): If (x + α) is a factor of f(x), then f(α) = 0.
Let,
⇒ f(x) = x3 + 2x2 - x - 2
⇒ f(1) = (1)3 + 2(1)2 - 1 - 2
= 1 + 2 - 1 - 2
= 3 - 3
= 0.
Since, f(1) = 0.
Thus, (x − 1) is a factor of f(x) = x3 + 2x2 - x - 2 if f(1) = 0.
∴ Assertion (A) is true.
⇒ x + a = 0
⇒ x = -a.
If (x + a) is a factor of f(x), then f(−a) = 0.
∴ Reason (R) is false.
A is true, R is false.
Hence, option 3 is the correct option.
Assertion (A): If (2x - 1) is a factor of polynomial f(x), then $f\Big(\frac{1}{2}\Big)$ = 0.
Reason (R): (ax + b) is a factor of f(x) implies $f\Big(-\frac{b}{a}\Big)$ = 0.
By factor theorem,
If (x - a) is a factor of f(x), then f(a) = 0.
Given,
⇒ 2x - 1 = 0
⇒ 2x = 1
⇒ x = $\frac{1}{2}$
Thus, if (2x - 1) is a factor of polynomial f(x), then $f\Big(\frac{1}{2}\Big)$ = 0.
∴ Assertion (A) is true.
⇒ ax + b = 0
⇒ ax = -b
⇒ x = $-\frac{b}{a}$
Thus, if (ax + b) is a factor of polynomial f(x), then $f\Big(-\frac{b}{a}\Big)$ = 0.
∴ Reason (R) is true.
Thus, both A and R are true, and R is the correct explanation of A.
Hence, option 1 is the correct option.
Assertion (A): x3 + 2x2 - x - 2 is a polynomial of degree 3.
Reason (R): x + 2 is a factor of the polynomial.
Since, the highest value of power of x in the polynomial x3 + 2x2 - x - 2 is 3.
Thus,
x3 + 2x2 - x - 2 is a polynomial of degree 3.
∴ Assertion (A) is true.
By factor theorem,
(x - a) is a factor of f(x) if f(a) = 0.
x + 2 = 0
x = -2
Substituting x = -2 in x3 + 2x2 - x - 2, we get :
⇒ (-2)3 + 2(-2)2 - (-2) - 2
⇒ -8 + 2(4) + 2 - 2
⇒ -8 + 8 + 2 - 2
⇒ 0.
∴ Reason (R) is true.
Both A and R are true, but R is not the correct explanation of A.
Hence, Option 2 is the correct option.
Pamela factorized the following polynomial :
2x3 + 3x2 - 3x - 2
She found the result as (x + 2)(x - 1)(x - 2). Using remainder and factor theorem, verify whether her result is correct. If incorrect, give the correct result.
Dividing the polynomial f(x) = 2x3 + 3x2 - 3x - 2 by (x + 2), we get :
⇒ x + 2 = 0
⇒ x = -2
f(-2) = 2(-2)3 + 3(-2)2 - 3(-2) - 2
= 2(-8) + 3(4) + 6 - 2
= -16 + 12 + 6 - 2
= -18 + 18
= 0.
Dividing the polynomial f(x) = 2x3 + 3x2 - 3x - 2 by (x - 1), we get :
⇒ x - 1 = 0
⇒ x = 1
f(1) = 2(1)3 + 3(1)2 - 3(1) - 2
= 2(1) + 3(1) - 3 - 2
= 2 + 3 - 3 - 2
= 0.
Dividing the polynomial f(x) = 2x3 + 3x2 - 3x - 2 by (x - 2), we get :
⇒ x - 2 = 0
⇒ x = 2
f(2) = 2(2)3 + 3(2)2 - 3(2) - 2
= 2(8) + 3(4) - 6 - 2
= 16 + 12 - 6 - 2
= 20.
Since, f(2) ≠ 0
∴ x - 2 does not divide the polynomial 2x3 + 3x2 - 3x - 2.
Dividing the polynomial f(x) by (x + 2)(x - 1) or by (x2 + x - 2), we get :
$$\begin{array}{l} \phantom{x^2 + x - 2)}{\quad 2x + 1} \\ x^2 + x - 2\overline{\smash{\big)}\quad 2x^3 + 3x^2 - 3x - 2} \\ \phantom{x^2 + x - 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{-}{+}2x^2 \underset{+}{-}4x} \\ \phantom{{x^2 + x - 2}2x^3 + 20x^3}x^2 + x - 2 \\ \phantom{{x^2 + x - 2)}2x^3 + x^3}\underline{\underset{-}{+}x^2 \underset{-}{+} x \underset{+}{-} 2} \\ \phantom{{x^2 + x - 2)}{x^3-2x^{2}(31)}{x}}\times \end{array}$$
∴ 2x3 + 3x2 - 3x - 2 = (x2 + x - 2)(2x + 1)
= (x + 2)(x - 1)(2x + 1).
Hence, 2x3 + 3x2 - 3x - 2 = (x + 2)(x - 1)(2x + 1).
Using remainder and factor theorem, show that (2x + 3) is a factor of the polynomial 2x2 + 11x + 12. Hence, factorise it completely. What must be multiplied to the given polynomial so that x2 + 3x - 4 is a factor of the resulting polynomial? Also, write the resulting polynomial.
2x + 3 = 0
⇒ 2x = -3
⇒ x = $-\frac{3}{2}$
Substituting value of x in equation 2x2 + 11x + 12, we get :
$$\begin{aligned}\Rightarrow 2 \times \Big(-\frac{3}{2}\Big)^2 + 11 \times \Big(-\frac{3}{2}\Big) + 12 \\[1em] \Rightarrow 2 \times \frac{9}{4} - \frac{33}{2} + 12 \\[1em] \Rightarrow \frac{9}{2} - \frac{33}{2} + 12 \\[1em] \Rightarrow \frac{9 - 33 + 24}{2} \\[1em] \Rightarrow \frac{0}{2} \\[1em] \Rightarrow 0.\end{aligned}$$
Since, remainder = 0.
∴ 2x + 3 is a factor of the polynomial 2x2 + 11x + 12.
Solving polynomial, 2x2 + 11x + 12, we get :
⇒ 2x2 + 8x + 3x + 12
⇒ 2x(x + 4) + 3(x + 4)
⇒ (2x + 3)(x + 4).
Solving polynomial, x2 + 3x - 4, we get :
⇒ x2 + 4x - x - 4
⇒ x(x + 4) - 1(x + 4)
⇒ (x - 1)(x + 4).
∴ (x - 1) and (x + 4) are factors of x2 + 3x - 4.
∴ On multiplying polynomial, 2x2 + 11x + 12 by (x - 1) it will be divisible by x2 + 3x - 4.
⇒ (2x2 + 11x + 12)(x - 1)
⇒ 2x3 - 2x2 + 11x2 - 11x + 12x - 12
⇒ 2x3 + 9x2 + x - 12.
Hence, the resulting polynomial = 2x3 + 9x2 + x - 12.
Given, 9𝑥2 - 4 is a factor of 9𝑥3 - m𝑥2 - n𝑥 + 8 :
(i) find the value of m and n using the remainder and factor theorem.
(ii) factorise the given polynomial completely.
(i) Simplifying 9𝑥2 - 4, we get :
⇒ 9𝑥2 - 4
⇒ (3𝑥)2 - 22
⇒ (3𝑥 + 2)(3𝑥 - 2).
We know that,
If (x - a) is a factor of f(x), then f(a) = 0.
Given,
9𝑥2 - 4 is a factor of 9𝑥3 - m𝑥2 - n𝑥 + 8.
∴ (3𝑥 + 2) and (3𝑥 - 2) are the factors of 9𝑥3 - m𝑥2 - n𝑥 + 8.
⇒ 3𝑥 + 2 = 0
⇒ 3𝑥 = -2
⇒ 𝑥 = $-\frac{2}{3}$
Substituting x = $-\frac{2}{3}$ in 9𝑥3 - mx2 - nx + 8, we get remainder = 0.
$$\begin{aligned}\Rightarrow 9 \times \Big(-\frac{2}{3}\Big)^3 - m \times \Big(-\frac{2}{3}\Big)^2 - n \times \Big(-\frac{2}{3}\Big) + 8 = 0 \\[1em] \Rightarrow 9 \times -\frac{8}{27} - m \times \frac{4}{9} + \frac{2n}{3} + 8 = 0 \\[1em] \Rightarrow -\frac{8}{3} - \frac{4m}{9} + \frac{2n}{3} + 8 = 0 \\[1em] \Rightarrow \frac{-24 - 4m + 6n + 72}{9} = 0 \\[1em] \Rightarrow \frac{6n - 4m + 48}{9} = 0 \\[1em] \Rightarrow 6n - 4m + 48 = 0 \\[1em] \Rightarrow 2(3n - 2m + 24) = 0 \\[1em] \Rightarrow 3n - 2m + 24 = 0 \\[1em] \Rightarrow 3n - 2m = -24 \text{ ..........(1)}\end{aligned}$$
⇒ 3𝑥 - 2 = 0
⇒ 3𝑥 = 2
⇒ 𝑥 = $\frac{2}{3}$
Substituting 𝑥 = $\frac{2}{3}$ in 9𝑥3 - m𝑥2 - n𝑥 + 8, we get remainder = 0.
$$\begin{aligned}\Rightarrow 9 \times \Big(\frac{2}{3}\Big)^3 - m \times \Big(\frac{2}{3}\Big)^2 - n \times \Big(\frac{2}{3}\Big) + 8 = 0 \\[1em] \Rightarrow 9 \times \frac{8}{27} - m \times \frac{4}{9} - \frac{2n}{3} + 8 = 0 \\[1em] \Rightarrow \frac{8}{3} - \frac{4m}{9} - \frac{2n}{3} + 8 = 0 \\[1em] \Rightarrow \frac{24 - 4m - 6n + 72}{9} = 0 \\[1em] \Rightarrow \frac{96 - 6n - 4m}{9} = 0 \\[1em] \Rightarrow 96 - 6n - 4m = 0 \\[1em] \Rightarrow 4m + 6n = 96 \\[1em] \Rightarrow 2(2m + 3n) = 96 \\[1em] \Rightarrow 2m + 3n = \frac{96}{2} \\[1em] \Rightarrow 2m + 3n = 48 \text{ ..........(2)}\end{aligned}$$
Adding equation (1) and (2), we get :
⇒ 3n - 2m + 2m + 3n = -24 + 48
⇒ 6n = 24
⇒ n = $\frac{24}{6}$
⇒ n = 4.
Substituting value of n in equation (1), we get :
⇒ 3(4) - 2m = -24
⇒ 12 - 2m = -24
⇒ -2m = -24 - 12
⇒ -2m = -36
⇒ m = $\frac{-36}{-2}$ = 18.
Hence, m = 18 and n = 4.
(ii) Substituting value of m and n in 9𝑥3 - m𝑥2 - n𝑥 + 8, we get :
9𝑥3 - 18𝑥2 - 4𝑥 + 8
Dividing 9𝑥3 - 18𝑥2 - 4𝑥 + 8 by 9𝑥2 - 4, we get :
$$\begin{array}{l} \phantom{9x^2 - 4)}{\quad x - 2} \\ 9x^2 - 4\overline{\smash{\big)}\quad 9x^3 - 18x^2 - 4x + 8} \\ \phantom{9x^2 - 4)}\phantom{)}\underline{\underset{-}{+}9x^3 \phantom{- 18x^2} \underset{+}{-}4x} \\ \phantom{{9x^2 - 4}{+9x^3 - }}-18x^2 \phantom{- 4x} + 8 \\ \phantom{{9x^2 - 4)}{+9x^3 - }}\underline{\underset{+}{-}18x^2 \phantom{- 4x} \underset{-}{+} 8} \\ \phantom{{9x^2 - 4)}{x^3-2x^{2}(31)}{x}}\times \end{array}$$
∴ 9𝑥3 - 18𝑥2 - 4𝑥 + 8 = (9𝑥2 - 4)(𝑥 - 2)
= (3𝑥 + 2)(3𝑥 - 2)(𝑥 - 2).
Hence, factors are (3𝑥 + 2), (3𝑥 - 2) and (𝑥 - 2).
If a polynomial x3 + 2x2 – ax + b leaves a remainder -6 when divided by x + 1 and the same polynomial has x - 2 as a factor, then find the values of a and b.
Given,
x3 + 2x2 – ax + b leaves a remainder -6 when divided by x + 1.
⇒ x + 1 = 0
⇒ x = -1.
Let f(x) = x3 + 2x2 – ax + b, then f(-1) = -6.
⇒ f(-1) = -6
⇒ (-1)3 + 2(-1)2 - a(-1) + b = -6
⇒ -1 + 2(1) + a + b = -6
⇒ -1 + 2 + a + b = -6
⇒ a + b + 1 = -6
⇒ a + b = -6 - 1
⇒ a + b = -7 ........(1)
Given ,
x - 2 is a factor of f(x).
⇒ x - 2 = 0
⇒ x = 2.
f(2) = 0
⇒ 23 + 2(2)2 – 2a + b = 0
⇒ 8 + 2(4) - 2a + b = 0
⇒ 8 + 8 - 2a + b = 0
⇒ 16 - 2a + b = 0
⇒ 2a - b = 16 ......(2)
Adding equation (1) and (2), we get :
⇒ a + b + 2a - b = -7 + 16
⇒ 3a = 9
⇒ a = $\frac{9}{3}$ = 3.
Substituting value of a = 3 in equation (1), we get :
⇒ 3 + b = -7
⇒ b = -7 - 3 = -10.
Hence, a = 3 and b = -10.
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