ICSE Class 10 Mathematics · Selina Concise Mathematics · 56 solved questions
x - 1 is a factor of 8x2 - 7x + m; the value of m is :
By factor theorem,
If x - a is a factor of polynomial f(x), then remainder f(a) = 0.
Given,
x - 1 is a factor of 8x2 - 7x + m.
⇒ x - 1 = 0
⇒ x = 1.
Substituting x = 1, in 8x2 - 7x + m remainder will be zero.
⇒ 8(1)2 - 7(1) + m = 0
⇒ 8 - 7 + m = 0
⇒ 1 + m = 0
⇒ m = -1.
Hence, Option 1 is the correct option.
If (x - a) is a factor of f(x) then the remainder when f(x) - k is divided by (x - a) is :
By factor theorem,
If x - a is a factor of polynomial f(x), then remainder f(a) = 0.
Given,
(x - a) is a factor of f(x), then f(a) = 0.
Substituting x = a, in f(x) - k,
⇒ f(a) - k
⇒ 0 - k
⇒ -k.
Hence, Option 3 is the correct option.
One factor of x3 - kx2 + 11x - 6 is x - 1. The value of k is :
By factor theorem,
If x - a is a factor of polynomial f(x), then remainder f(a) = 0.
Given,
x - 1 is a factor of 8x2 - 7x + m.
⇒ x - 1 = 0
⇒ x = 1.
Substituting x = 1, in x3 - kx2 + 11x - 6 remainder will be zero.
⇒ 13 - k(1)2 + 11(1) - 6 = 0
⇒ 1 - k + 11 - 6 = 0
⇒ 6 - k = 0
⇒ k = 6.
Hence, Option 3 is the correct option.
If (x - a) is a factor of x3 - ax2 + x + 5; the value of a is :
By factor theorem,
If polynomial ƒ(x) is divided by its factor (x - a) then the remainder ƒ(a) = 0.
Since, x - a is a factor of x3 - ax2 + x + 5.
∴ On substituting x = a in x3 - ax2 + x + 5, remainder = 0.
∴ a3 - a(a)2 + a + 5 = 0
⇒ a3 - a3 + a + 5 = 0
⇒ a + 5 = 0
⇒ a = -5.
Hence, Option 4 is the correct option.
(x - 2) is a factor of :
⇒ x - 2 = 0
⇒ x = 2.
Substituting x = 2 in x3 - x2 + x - 6, we get :
⇒ 23 - 22 + 2 - 6
⇒ 8 - 4 + 2 - 6
⇒ 10 - 10
⇒ 0.
Since, remainder = 0.
∴ x - 2 is a factor of x3 - x2 + x - 6.
Hence, Option 1 is the correct option.
Find in each case, the remainder when :
x4 - 3x2 + 2x + 1 is divided by x - 1
x - 1 = 0 ⇒ x = 1.
Required remainder = Value of given polynomial x4 - 3x2 + 2x + 1 at x = 1.
∴ Remainder = (1)4 - 3(1)2 + 2(1) + 1
= 1 - 3 + 2 + 1
= 1.
Hence, remainder = 1.
Find in each case, the remainder when :
x3 + 3x2 - 12x + 4 is divided by x - 2.
x - 2 = 0 ⇒ x = 2.
Required remainder = Value of given polynomial x3 + 3x2 - 12x + 4 at x = 2.
∴ Remainder = (2)3 + 3(2)2 - 12(2) + 4
= 8 + 12 - 24 + 4
= 0.
Hence, remainder = 0.
Find in each case, the remainder when :
x4 + 1 is divisible by x + 1.
x + 1 = 0 ⇒ x = -1.
Required remainder = Value of given polynomial x4 + 1 at x = -1.
∴ Remainder = (-1)4 + 1
= 1 + 1
= 2.
Hence, remainder = 2.
Use the Remainder theorem to find which of the following is a factor of 2x3 + 3x2 - 5x - 6.
(i) x + 1
(ii) 2x - 1
(i) x + 1 = 0 ⇒ x = -1
Required remainder = Value of given polynomial 2x3 + 3x2 - 5x - 6 at x = -1.
∴ Remainder = 2(-1)3 + 3(-1)2 - 5(-1) - 6
= 2(-1) + 3(1) + 5 - 6
= -2 + 3 + 5 - 6
= 8 - 8
Since, remainder = 0
∴ x + 1 is a factor of 2x3 + 3x2 - 5x - 6
(ii) 2x - 1 = 0 ⇒ x = $\frac{1}{2}$
Required remainder = Value of given polynomial 2x3 + 3x2 - 5x - 6 at x = $\frac{1}{2}$.
$$\begin{aligned}\therefore \text{ Remainder} = 2\Big(\frac{1}{2}\Big)^3 + 3\Big(\frac{1}{2}\Big)^2 - 5\Big(\frac{1}{2}\Big) - 6 \\[1em] = 2 \times \frac{1}{8} + 3 \times \frac{1}{4} - \frac{5}{2} - 6 \\[1em] = \frac{1}{4} + \frac{3}{4} - \frac{5}{2} - 6 \\[1em] = \frac{1 + 3 - 10 - 24}{4} \\[1em] = -\frac{30}{4} \\[1em] = -\frac{15}{2}.\end{aligned}$$
Since, remainder ≠ 0
∴ 2x - 1 is not a factor of 2x3 + 3x2 - 5x - 6.
If 2x + 1 is a factor of 2x2 + ax - 3, find the value of a.
2x + 1 = 0 ⇒ x = $-\frac{1}{2}$
Since, 2x + 1 is a factor of 2x2 + ax - 3
∴ On substituting x = $-\frac{1}{2}$ in 2x2 + ax - 3 remainder = 0.
$$\begin{aligned}\Rightarrow 2\Big(-\frac{1}{2}\Big)^2 + a\Big(-\frac{1}{2}\Big) - 3 = 0 \\[1em] \Rightarrow 2 \times \frac{1}{4} - \frac{a}{2} - 3 = 0 \\[1em] \Rightarrow \frac{1}{2} - \frac{a}{2} = 3 \\[1em] \Rightarrow \frac{1 - a}{2} = 3 \\[1em] \Rightarrow 1 - a = 6 \\[1em] \Rightarrow a = 1 - 6 = -5.\end{aligned}$$
Hence, a = -5.
Find the values of constants a and b when x - 2 and x + 3 both are the factors of expression x3 + ax2 + bx - 12.
x - 2 = 0 ⇒ x = 2
Since, x - 2 is a factor of x3 + ax2 + bx - 12
∴ On substituting x = 2 in x3 + ax2 + bx - 12, remainder = 0.
⇒ (2)3 + a(2)2 + b(2) - 12 = 0
⇒ 8 + 4a + 2b - 12 = 0
⇒ 4a + 2b - 4 = 0
⇒ 4a + 2b = 4
⇒ 2(2a + b) = 4
⇒ 2a + b = 2
⇒ b = 2 - 2a .........(i)
x + 3 = 0 ⇒ x = -3
Since, x + 3 is a factor of x3 + ax2 + bx - 12
∴ On substituting x = -3 in x3 + ax2 + bx - 12, remainder = 0.
⇒ (-3)3 + a(-3)2 + b(-3) - 12 = 0
⇒ -27 + 9a - 3b - 12 = 0
⇒ 9a - 3b - 39 = 0
⇒ 9a - 3b = 39
⇒ 3(3a - b) = 39
⇒ 3a - b = 13
⇒ b = 3a - 13 .........(ii)
From (i) and (ii) we get,
⇒ 2 - 2a = 3a - 13
⇒ 3a + 2a = 2 + 13
⇒ 5a = 15
⇒ a = 3.
Substituting value of a in (i),
⇒ b = 2 - 2a = 2 - 2(3) = 2 - 6 = -4.
Hence, a = 3 and b = -4.
Find the value of k, if 2x + 1 is a factor of (3k + 2)x3 + (k - 1).
2x + 1 = 0 ⇒ x = -$\frac{1}{2}$
Since, 2x + 1 is a factor of (3k + 2)x3 + (k - 1)
∴ On substituting x = $-\frac{1}{2}$ in (3k + 2)x3 + (k - 1), remainder = 0.
$$\begin{aligned}\Rightarrow (3k + 2)\Big(-\frac{1}{2}\Big)^3 + (k - 1) = 0 \\[1em] \Rightarrow \frac{-(3k + 2)}{8} + (k - 1) = 0 \\[1em] \Rightarrow \frac{-3k - 2 + 8k - 8}{8} = 0 \\[1em] \Rightarrow 5k - 10 = 0 \\[1em] \Rightarrow 5k = 10 \\[1em] \Rightarrow k = 2.\end{aligned}$$
Hence, k = 2.
Find the values of m and n so that x - 1 and x + 2 both are factors of
x3 + (3m + 1)x2 + nx - 18.
x - 1 = 0 ⇒ x = 1.
Since, x - 1 is a factor of x3 + (3m + 1)x2 + nx - 18,
∴ On substituting x = 1 in x3 + (3m + 1)x2 + nx - 18, remainder = 0.
⇒ (1)3 + (3m + 1)(1)2 + n(1) - 18 = 0
⇒ 1 + 3m + 1 + n - 18 = 0
⇒ 3m + n - 16 = 0
⇒ n = 16 - 3m .........(i)
x + 2 = 0 ⇒ x = -2.
Since, x + 2 is a factor of x3 + (3m + 1)x2 + nx - 18,
∴ On substituting x = -2 in x3 + (3m + 1)x2 + nx - 18, remainder = 0.
(-2)3 + (3m + 1)(-2)2 + n(-2) - 18 = 0
⇒ -8 + (3m + 1)(4) - 2n - 18 = 0
⇒ -8 + 12m + 4 - 2n - 18 = 0
⇒ 12m - 2n - 22 = 0
⇒ 12m - 2n = 22
⇒ 2(6m - n) = 22
⇒ 6m - n = 11
⇒ n = 6m - 11 .........(ii)
From (i) and (ii) we get,
⇒ 16 - 3m = 6m - 11
⇒ 6m + 3m = 16 + 11
⇒ 9m = 27
⇒ m = 3.
Substituting m = 3 in (ii) we get,
⇒ n = 6(3) - 11 = 18 - 11 = 7.
Hence, m = 3 and n = 7.
When x3 + 2x2 - kx + 4 is divided by x - 2, the remainder is k. Find the value of constant k.
x - 2 = 0 ⇒ x = 2.
Given, when x3 + 2x2 - kx + 4 is divided by x - 2, the remainder is k.
∴ On substituting x = 2 in x3 + 2x2 - kx + 4, remainder = k.
⇒ (2)3 + 2(2)2 - k(2) + 4 = k
⇒ 8 + 8 - 2k + 4 = k
⇒ 20 - 2k = k
⇒ 3k = 20
⇒ k = $\frac{20}{3} = 6\frac{2}{3}.$
Hence, k = $6\frac{2}{3}.$
Find the value of a, if the division of ax3 + 9x2 + 4x - 10 by x + 3 leaves a remainder 5.
x + 3 = 0 ⇒ x = -3.
Given, when ax3 + 9x2 + 4x - 10 is divided by x + 3, the remainder is 5.
∴ On substituting x = -3 in ax3 + 9x2 + 4x - 10, remainder = 5.
⇒ a(-3)3 + 9(-3)2 + 4(-3) - 10 = 5
⇒ -27a + 81 - 12 - 10 = 5
⇒ -27a + 59 = 5
⇒ 27a = 59 - 5
⇒ 27a = 54
⇒ a = $\frac{54}{27}$ = 2
Hence, a = 2.
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.
x - 2 = 0 ⇒ x = 2.
Since, x - 2 is a factor of x3 + ax2 + bx + 6,
∴ On substituting x = 2 in x3 + ax2 + bx + 6, remainder = 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
⇒ b = -(7 + 2a) .......(i)
x - 3 = 0 ⇒ x = 3.
Given, when x3 + ax2 + bx + 6 is divided by x - 3, the remainder is 3.
∴ On substituting x = 3 in x3 + ax2 + bx + 6, remainder = 3.
⇒ (3)3 + a(3)2 + b(3) + 6 = 3
⇒ 27 + 9a + 3b + 6 = 3
⇒ 9a + 3b + 33 = 3
⇒ 9a + 3b = -30
⇒ 3(3a + b) = -30
⇒ 3a + b = -10
⇒ b = -10 - 3a = -(10 + 3a) ........(ii)
From (i) and (ii) we get,
⇒ -(7 + 2a) = -(10 + 3a)
⇒ 7 + 2a = 10 + 3a
⇒ 3a - 2a = 7 - 10
⇒ a = -3.
Substituting a = -3 in (i) we get,
⇒ b = -(7 + 2a) = -(7 + 2(-3)) = -(7 - 6) = -1.
Hence, a = -3 and b = -1.
What number should be added to 3x3 - 5x2 + 6x so that when resulting polynomial is divided by x - 3, the remainder is 8 ?
Let number to be added be a.
∴ Polynomial = 3x3 - 5x2 + 6x + a
x - 3 = 0 ⇒ x = 3
On substituting x = 3 in 3x3 - 5x2 + 6x + a, remainder = 8.
∴ 3(3)3 - 5(3)2 + 6(3) + a = 8
⇒ 3(27) - 5(9) + 18 + a = 8
⇒ 81 - 45 + 18 + a = 8
⇒ a + 54 = 8
⇒ a = -46
Hence, no. to be added = -46.
What number should be subtracted from x3 + 3x2 - 8x + 14 so that on dividing it by x - 2, the remainder is 10 ?
Let number to be subtracted be a.
∴ Polynomial = x3 + 3x2 - 8x + 14 - a
x - 2 = 0 ⇒ x = 2
On substituting x = 2 in x3 + 3x2 - 8x + 14 - a, remainder = 10.
∴ (2)3 + 3(2)2 - 8(2) + 14 - a = 10
⇒ 8 + 3(4) - 16 + 14 - a = 10
⇒ 8 + 12 - 16 + 14 - a = 10
⇒ 18 - a = 10
⇒ a = 18 - 10 = 8.
Hence, no. to be subtracted = 8.
The polynomial 2x3 - 7x2 + ax - 6 and x3 - 8x2 + (2a + 1)x - 16 leave the same remainder when divided by x - 2. Find the value of 'a'.
Given,
2x3 - 7x2 + ax - 6 and x3 - 8x2 + (2a + 1)x - 16 leave the same remainder when divided by x - 2.
x - 2 = 0 ⇒ x = 2
∴ On substituting x = 2 in 2x3 - 7x2 + ax - 6 and x3 - 8x2 + (2a + 1)x - 16 the values are equal.
∴ 2(2)3 - 7(2)2 + a(2) - 6 = (2)3 - 8(2)2 + (2a + 1)(2) - 16
⇒ 2(8) - 7(4) + 2a - 6 = 8 - 32 + 4a + 2 - 16
⇒ 16 - 28 + 2a - 6 = 8 - 32 + 4a + 2 - 16
⇒ 2a - 18 = 4a - 38
⇒ 4a - 2a = 38 - 18
⇒ 2a = 20
⇒ a = 10.
Hence, a = 10.
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.
Given,
(x - 2) is a factor of the expression 2x3 + ax2 + bx - 14.
x - 2 = 0 ⇒ x = 2
∴ On substituting x = 2 in 2x3 + ax2 + bx - 14, remainder = 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
⇒ 2(2a + b + 1) = 0
⇒ 2a + b + 1 = 0
⇒ b = -(1 + 2a) .......(i)
Given,
On dividing 2x3 + ax2 + bx - 14 by (x - 3), remainder = 52
x - 3 = 0 ⇒ x = 3
∴ On substituting x = 3 in 2x3 + ax2 + bx - 14, remainder = 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 = 12
⇒ 3(3a + b) = 12
⇒ 3a + b = 4
⇒ b = 4 - 3a ........(ii)
From (i) and (ii) we get,
⇒ -(1 + 2a) = 4 - 3a
⇒ -1 - 2a = 4 - 3a
⇒ -2a + 3a = 4 + 1
⇒ a = 5.
Substituting value of a in (i) we get,
⇒ b = -(1 + 2a) = -(1 + 2(5)) = -(1 + 10) = -11.
Hence, a = 5 and b = -11.
For the polynomial x5 - x4 + x3 - 8x2 + 6x + 15, the maximum number of linear factors is :
Maximum number of linear factors in a polynomial depends upon the highest power of variable.
Maximum number of linear factors in x5 - x4 + x3 - 8x2 + 6x + 15 = 5.
Hence, Option 4 is the correct option.
If f(x) = 3x + 8; the value of f(x) + f(-x) is :
Given,
f(x) = 3x + 8
f(-x) = 3(-x) + 8 = -3x + 8
f(x) + f(-x) = 3x + 8 + (-3x + 8)
= 3x - 3x + 8 + 8
= 16.
Hence, Option 2 is the correct option.
If x25 + x24 is divided by (x + 1), the result is :
By remainder theorem,
The remainder theorem states that when a polynomial f(x) is divided by (x - a), then the remainder = f(a).
Given,
⇒ x + 1 = 0
⇒ x = -1.
Substituting x = -1 in x25 + x24, we get :
⇒ (-1)25 + (-1)24
⇒ -1 + 1
⇒ 0.
Hence, Option 3 is the correct option.
Factors of 3x3 - 2x2 - 8x are :
Given,
⇒ 3x3 - 2x2 - 8x
⇒ x[3x2 - 2x - 8]
⇒ x[3x2 - 6x + 4x - 8]
⇒ x[3x(x - 2) + 4(x - 2)]
⇒ x[(3x + 4)(x - 2)]
⇒ x(3x + 4)(x - 2).
Hence, Option 2 is the correct option.
Factors of 4 + 4x - x2 - x3 are :
Substituting x = 2 in 4 + 4x - x2 - x3, we get :
⇒ 4 + 4x - x2 - x3
⇒ 4 + 4(2) - 22 - 23
⇒ 4 + 8 - 4 - 8
⇒ 0.
∴ (x - 2) is the factor of 4 + 4x - x2 - x3.
Dividing -x3 - x2 + 4x + 4 by (x - 2),
$$\begin{array}{l} \phantom{x - 2)}{-x^2 -3x - 2} \\ x - 2\overline{\smash{\big)}-x^3 - x^2 + 4x + 4} \\ \phantom{x - 2)}\phantom{2}\underline{\underset{+}{-}x^3 \underset{-}{+}2x^2} \\ \phantom{{x - 2}x^3-2}-3x^2 + 4x \\ \phantom{{x - 2)}x^3-2}\underline{\underset{+}{-}3x^2 \underset{-}{+} 6x} \\ \phantom{{x - 2)}{x^3-2x^{2}(3)}}-2x + 4 \\ \phantom{{x - 2)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}2x \underset{-}{+} 4} \\ \phantom{{x - 2)}{x^3-2x^{2}(31)}{-2x}}\times \end{array}$$
we get quotient = -x2 - 3x - 2.
∴ -x3 - x2 + 4x + 4 = (x - 2)(-x2 - 3x - 2)
= (x - 2)[-x2 - 2x - x - 2]
= (x - 2)[-x(x + 2) - 1(x + 2)]
= (x - 2)(x + 2)(-x - 1)
= -(x - 2)(x + 2)(x + 1)
= (2 - x)(x + 2)(x + 1)
Rearranging the terms we get,
⇒ (2 + x)(2 - x)(1 + x)
Hence, Option 1 is the correct option.
Using Factor Theorem, show that :
(x - 2) is a factor of x3 - 2x2 - 9x + 18. Hence, factorise the expression x3 - 2x2 - 9x + 18 completely.
x - 2 = 0 ⇒ x = 2.
Remainder = The value of x3 - 2x2 - 9x + 18 at x = 2.
= (2)3 - 2(2)2 - 9(2) + 18
= 8 - 8 - 18 + 18
= 0.
Hence, (x - 2) is a factor of x3 - 2x2 - 9x + 18.
Now dividing x3 - 2x2 - 9x + 18 by (x - 2),
$$\begin{array}{l} \phantom{x - 2)}{x^2 - 9} \\ x - 2\overline{\smash{\big)}x^3 - 2x^2 - 9x + 18} \\ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}2x^2} \\ \phantom{{x - 2}x^3-2x^2}-9x + 18 \\ \phantom{{x - 2}x^3-2x^2\space}\underline{\underset{+}{-}9x \underset{-}{+} 18} \\ \phantom{{x - 2}{x^3-2x^2\space}{-9x}}\times \end{array}$$
we get quotient = x2 - 9
∴ x3 - 2x2 - 9x + 18 = (x - 2)(x2 - 9) = (x - 2)(x - 3)(x + 3).
Hence, x3 - 2x2 - 9x + 18 = (x - 2)(x - 3)(x + 3).
Using Factor Theorem, show that :
(x + 5) is a factor of 2x3 + 5x2 - 28x - 15. Hence, factorise the expression 2x3 + 5x2 - 28x - 15 completely.
x + 5 = 0 ⇒ x = -5.
Remainder = The value of 2x3 + 5x2 - 28x - 15 at x = -5.
= 2(-5)3 + 5(-5)2 - 28(-5) - 15
= 2(-125) + 5(25) + 140 - 15
= -250 + 125 + 140 - 15
= -265 + 265
= 0.
Hence, (x + 5) is a factor of 2x3 + 5x2 - 28x - 15.
Now dividing 2x3 + 5x2 - 28x - 15 by (x + 5),
$$\begin{array}{l} \phantom{x + 5)}{2x^2 - 5x - 3} \\ x + 5\overline{\smash{\big)}2x^3 + 5x^2 - 28x - 15} \\ \phantom{x - 5}\underline{\underset{-}{}2x^3 \underset{-}{+}10x^2} \\ \phantom{{x - 5}2x^3+}-5x^2 - 28x \\ \phantom{{x - 5}2x^3+}\underline{\underset{+}{-}5x^2 \underset{+}{-} 25x} \\ \phantom{{x - 5}{2x^3+}{+5x^2+}}-3x - 15 \\ \phantom{{x - 5}{2x^3+}{+5x^2+\enspace}}\underline{\underset{+}{-}3x \underset{+}{-} 15} \\ \phantom{{x - 5}{2x^3+}{+5x^2+\enspace}{-3x}}\times \end{array}$$
we get quotient = 2x2 - 5x - 3
Factorising 2x2 - 5x - 3,
⇒ 2x2 - 6x + x - 3
⇒ 2x(x - 3) + 1(x - 3)
⇒ (2x + 1)(x - 3)
∴ 2x3 + 5x2 - 28x - 15 = (x + 5)(2x + 1)(x - 3).
Hence, 2x3 + 5x2 - 28x - 15 = (x + 5)(2x + 1)(x - 3).
Using the remainder theorem, factorise each of the following completely :
3x3 + 2x2 - 19x + 6
For x = 2, the value of 3x3 + 2x2 - 19x + 6,
= 3(2)3 + 2(2)2 - 19(2) + 6
= 3(8) + 2(4) - 38 + 6
= 38 - 38
= 0.
Hence, (x - 2) is the factor of 3x3 + 2x2 - 19x + 6.
On dividing 3x3 + 2x2 - 19x + 6 by (x - 2),
$$\begin{array}{l} \phantom{x - 2)}{3x^2 + 8x - 3} \\ x - 2\overline{\smash{\big)}3x^3 + 2x^2 - 19x + 6} \\ \phantom{x - 2}\underline{\underset{-}{}3x^3 \underset{+}{-} 6x^2} \\ \phantom{{x - 2}2x^3+4}8x^2 - 19x \\ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}8x^2 \underset{-}{+} 16x} \\ \phantom{{x - 2}{2x^3+}{+2x^2}}-3x + 6 \\ \phantom{{x - 2}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}3x \underset{-}{+} 6} \\ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$$
we get, quotient = 3x2 + 8x - 3
Factorising, 3x2 + 8x - 3
= 3x2 + 9x - x - 3
= 3x(x + 3) - 1(x + 3)
= (3x - 1)(x + 3)
∴ 3x2 + 8x - 3 = (3x - 1)(x + 3).
Hence, 3x3 + 2x2 - 19x + 6 = (x - 2)(3x - 1)(x + 3).
Using the remainder theorem, factorise each of the following completely :
2x3 + x2 - 13x + 6
For x = 2, the value of 2x3 + x2 - 13x + 6
= 2(2)3 + (2)2 - 13(2) + 6
= 2(8) + 4 - 26 + 6
= 26 - 26
= 0.
On dividing 2x3 + x2 - 13x + 6 by (x - 2),
$$\begin{array}{l} \phantom{x - 2)}{2x^2 + 5x - 3} \\ x - 2\overline{\smash{\big)}2x^3 + x^2 - 13x + 6} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-} 4x^2} \\ \phantom{{x - 2}2x^3+4}5x^2 - 13x \\ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}5x^2 \underset{+}{-} 10x} \\ \phantom{{x - 2}{2x^3+}{+2x^2}}-3x + 6 \\ \phantom{{x - 2}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}3x \underset{-}{+} 6} \\ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$$
we get, quotient = 2x2 + 5x - 3
Factorising, 2x2 + 5x - 3
= 2x2 + 6x - x - 3
= 2x(x + 3) - 1(x + 3)
= (2x - 1)(x + 3)
∴ 2x2 + 5x - 3 = (2x - 1)(x + 3)
Hence, 2x3 + x2 - 13x + 6 = (x - 2)(2x - 1)(x + 3).
Using the remainder theorem, factorise each of the following completely :
3x3 + 2x2 - 23x - 30
For x = -2 the value of 3x3 + 2x2 - 23x - 30,
= 3(-2)3 + 2(-2)2 - 23(-2) - 30
= 3(-8) + 2(4) + 46 - 30
= -24 + 8 + 46 - 30
= 54 - 54
= 0.
Hence, (x + 2) is the factor of 3x3 + 2x2 - 23x - 30.
On dividing 3x3 + 2x2 - 23x - 30 by x + 2,
$$\begin{array}{l} \phantom{x + 2)}{3x^2 - 4x - 15} \\ x + 2\overline{\smash{\big)}3x^3 + 2x^2 - 23x - 30} \\ \phantom{x + 2}\underline{\underset{-}{}3x^3 \underset{-}{+} 6x^2} \\ \phantom{{x + 2}2x^3+}-4x^2 - 23x \\ \phantom{{x + 2}2x^3+}\underline{\underset{+}{-}4x^2 \underset{+}{-} 8x} \\ \phantom{{x + 2}{2x^3+}{+2x^2}}-15x - 30 \\ \phantom{{x + 2}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}15x \underset{+}{-} 30} \\ \phantom{{x + 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$$
we get, quotient = 3x2 - 4x - 15
Factorising, 3x2 - 4x - 15
= 3x2 - 9x + 5x - 15
= 3x(x- 3) + 5(x - 3)
= (3x + 5)(x - 3).
∴ 3x2 - 4x - 15 = (3x + 5)(x - 3).
Hence, 3x3 + 2x2 - 23x - 30 = (x + 2)(3x + 5)(x - 3).
Using the remainder theorem, factorise each of the following completely :
4x3 + 7x2 - 36x - 63
For x = -3 the value of 4x3 + 7x2 - 36x - 63
= 4(-3)3 + 7(-3)2 - 36(-3) - 63
= 4(-27) + 7(9) + 108 - 63
= -108 + 63 + 108 - 63
= 0.
Hence, (x + 3) is the factor of 4x3 + 7x2 - 36x - 63.
On dividing 4x3 + 7x2 - 36x - 63 by (x + 3),
$$\begin{array}{l} \phantom{x + 3)}{4x^2 - 5x - 21} \\ x + 3\overline{\smash{\big)}4x^3 + 7x^2 - 36x - 63} \\ \phantom{x + 3}\underline{\underset{-}{}4x^3 \underset{-}{+} 12x^2} \\ \phantom{{x + 3}2x^3+}-5x^2 - 36x \\ \phantom{{x + 3}2x^3+}\underline{\underset{+}{-}5x^2 \underset{+}{-} 15x} \\ \phantom{{x + 3}{2x^3+}{+2x^2}}-21x - 63 \\ \phantom{{x + 3}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}21x \underset{+}{-} 63} \\ \phantom{{x + 3}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$$
we get, quotient = 4x2 - 5x - 21
Factorising 4x2 - 5x - 21,
= 4x2 - 12x + 7x - 21
= 4x(x - 3) + 7(x - 3)
= (4x + 7)(x - 3).
∴ 4x2 - 5x - 21 = (4x + 7)(x - 3)
Hence, 4x3 + 7x2 - 36x - 63 = (x + 3)(4x + 7)(x - 3).
Using the remainder theorem, factorise each of the following completely :
x3 + x2 - 4x - 4
For x = -1 the value of x3 + x2 - 4x - 4
= (-1)3 + (-1)2 - 4(-1) - 4
= -1 + 1 + 4 - 4
= 0.
Hence, (x + 1) is the factor of x3 + x2 - 4x - 4.
On dividing x3 + x2 - 4x - 4 by (x + 1),
$$\begin{array}{l} \phantom{x + 1)}{x^2 - 4} \\ x + 1\overline{\smash{\big)}x^3 + x^2 - 4x - 4} \\ \phantom{x + 1}\underline{\underset{-}{}x^3 \underset{-}{+} x^2} \\ \phantom{{x + 1}x^3+x^2-}-4x - 4 \\ \phantom{{x + 1}x^3+x^2-}\underline{\underset{+}{-}4x \underset{+}{-} 4} \\ \phantom{{x + 1}x^3+x^2-4x\enspace} \times \end{array}$$
we get, quotient = x2 - 4
Factorising x2 - 4,
= (x)2 - 4
= (x + 2)(x - 2)
∴ x2 - 4 = (x - 2)(x + 2)
Hence, x3 + x2 - 4x - 4 = (x + 1)(x + 2)(x - 2).
Using the Remainder Theorem, factorise the expression 3x3 + 10x2 + x - 6. Hence, solve the equation 3x3 + 10x2 + x - 6 = 0.
For x = -1, the value of 3x3 + 10x2 + x - 6,
= 3(-1)3 + 10(-1)2 + (-1) - 6
= 3(-1) + 10(1) - 7
= 10 - 10
= 0.
Hence, (x + 1) is the factor of 3x3 + 10x2 + x - 6.
On dividing 3x3 + 10x2 + x - 6 by (x + 1),
$$\begin{array}{l} \phantom{x + 1)}{3x^2 + 7x - 6} \\ x + 1\overline{\smash{\big)}3x^3 + 10x^2 + x - 6} \\ \phantom{x + 1}\underline{\underset{-}{}3x^3 \underset{-}{+} 3x^2} \\ \phantom{{x + 1}3x^3+10}7x^2 + x \\ \phantom{{x + 1}3x^3+1}\underline{\underset{-}{}7x^2 \underset{-}{+} 7x} \\ \phantom{{x + 1}{3x^3+1}{+2x^2}}-6x - 6 \\ \phantom{{x + 1}{2x^3+}{+2x^2}{41\space}}\underline{\underset{+}{-}6x \underset{+}{-} 6} \\ \phantom{{x + 1}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$$
we get, quotient = 3x2 + 7x - 6
Factorising 3x2 + 7x - 6,
= 3x2 + 9x - 2x - 6
= 3x(x + 3) - 2(x + 3)
= (3x - 2)(x + 3).
∴ 3x2 + 7x - 6 = (3x - 2)(x + 3).
Hence, 3x3 + 10x2 + x - 6 = (x + 1)(3x - 2)(x + 3).
Factorise the expression
f(x) = 2x3 - 7x2 - 3x + 18.
Hence, find all possible values of x for which f(x) = 0.
For x = 2, the value of 2x3 - 7x2 - 3x + 18,
= 2(2)3 - 7(2)2 - 3(2) + 18
= 16 - 28 - 6 + 18
= 34 - 34
= 0.
Hence, (x - 2) is the factor of 2x3 - 7x2 - 3x + 18.
On dividing, 2x3 - 7x2 - 3x + 18 by (x - 2),
$$\begin{array}{l} \phantom{x - 2)}{2x^2 - 3x - 9} \\ x - 2\overline{\smash{\big)}2x^3 - 7x^2 - 3x + 18} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-} 4x^2} \\ \phantom{{x - 2}3x^3+}-3x^2 - 3x \\ \phantom{{x - 2}3x^3+}\underline{\underset{+}{-}3x^2 \underset{-}{+} 6x} \\ \phantom{{x - 2}{3x^3+1}{+2x^2}}-9x + 18 \\ \phantom{{x - 2}{2x^3+}{+2x^2}{41\space}}\underline{\underset{+}{-}9x \underset{-}{+} 18} \\ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$$
we get quotient = 2x2 - 3x - 9.
Factorising 2x2 - 3x - 9,
= 2x2 - 6x + 3x - 9
= 2x(x - 3) + 3(x - 3)
= (2x + 3)(x - 3).
∴ 2x2 - 3x - 9 = (2x + 3)(x - 3).
∴ f(x) = 2x3 - 7x2 - 3x + 18 = (x - 2)(2x + 3)(x - 3).
f(x) = 0, if (x - 2) = 0, (2x + 3) = 0 or x - 3 = 0.
x - 2 = 0 ⇒ x =2,
2x + 3 = 0 ⇒ x = -$\frac{3}{2}$,
x - 3 = 0 ⇒ x = 3.
Hence, 2x3 - 7x2 - 3x + 18 = (x - 2)(2x + 3)(x - 3), values for which f(x) = 0 are 2, 3, $-\frac{3}{2}$.
Given that x - 2 and x + 1 are factors of f(x) = x3 + 3x2 + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).
x - 2 = 0 ⇒ x = 2.
Since, x - 2 is a factor of x3 + 3x2 + ax + b. Hence, on substituting x = 2 in above expression, remainder = 0.
⇒ (2)3 + 3(2)2 + a(2) + b = 0
⇒ 8 + 12 + 2a + b = 0
⇒ 2a + b = -20
⇒ b = -20 - 2a ........(i)
x + 1 = 0 ⇒ x = -1.
Since, x + 1 is a factor of x3 + 3x2 + ax + b. Hence, on substituting x = -1 in above expression, remainder = 0.
⇒ (-1)3 + 3(-1)2 + a(-1) + b = 0
⇒ -1 + 3 - a + b = 0
⇒ 2 - a + b = 0
⇒ b = a - 2 .......(ii)
From (i) and (ii) we get,
⇒ -20 - 2a = a - 2
⇒ a + 2a = -20 + 2
⇒ 3a = -18
⇒ a = -6.
Substituting value of a in (ii) we get,
⇒ b = a - 2 = -6 - 2 = -8.
∴ a = -6 and b = -8.
On dividing, x3 + 3x2 - 6x - 8 by (x - 2),
$$\begin{array}{l} \phantom{x - 2)}{x^2 + 5x + 4} \\ x - 2\overline{\smash{\big)}x^3 + 3x^2 - 6x - 8} \\ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-} 2x^2} \\ \phantom{{x - 2}3x^3+}5x^2 - 6x \\ \phantom{{x - 2}3x^3\enspace\space}\underline{\underset{-}{}5x^2 \underset{+}{-} 10x} \\ \phantom{{x - 2}{3x^3+1}{+2x^2}}4x - 8 \\ \phantom{{x - 2}{2x^3+}{+2x^2}{4}}\underline{\underset{-}{}4x \underset{+}{-} 8} \\ \phantom{{x - 2}{2x^3+}{+2x^2-}{4x}}\times \end{array}$$
we get, quotient = x2 + 5x + 4.
Factorising x2 + 5x + 4,
= x2 + 4x + x + 4
= x(x + 4) + 1(x + 4)
= (x + 1)(x + 4).
Hence, a = -6, b = -8 and f(x) = (x - 2)(x + 1)(x + 4).
The expression 4x3 - bx2 + x - c leaves remainders 0 and 30 when divided by x + 1 and 2x - 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.
x + 1 = 0 ⇒ x = -1.
Since, x + 1 is a factor of 4x3 - bx2 + x - c. Hence, on substituting x = -1 in above expression, remainder = 0.
⇒ 4(-1)3 - b(-1)2 + (-1) - c = 0
⇒ -4 - b - 1 - c = 0
⇒ c = -5 - b .......(i)
Given, on dividing 4x3 - bx2 + x - c by (2x - 3) we get remainder = 30.
∴ On substituting x = $\frac{3}{2}$ in 4x3 - bx2 + x - c, value = 30.
$$\begin{aligned}\Rightarrow 4\Big(\frac{3}{2}\Big)^3 - b\Big(\frac{3}{2}\Big)^2 + \Big(\frac{3}{2}\Big) - c = 30 \\[1em] \Rightarrow 4\Big(\frac{27}{8}\Big) - b\Big(\frac{9}{4}\Big) + \Big(\frac{3}{2}\Big) - c = 30 \\[1em] \Rightarrow \frac{27}{2} - \frac{9b}{4} + \frac{3}{2} - 30 = c \\[1em] \Rightarrow \frac{54 - 9b + 6 - 120}{4} = c \\[1em] \Rightarrow \frac{-9b - 60}{4} = c ........(ii)\end{aligned}$$
From (i) and (ii) we get,
$$\begin{aligned}\Rightarrow -5 - b = \frac{-9b - 60}{4} \\[1em] \Rightarrow -20 - 4b = -9b - 60 \\[1em] \Rightarrow -20 + 60 = -9b + 4b \\[1em] \Rightarrow -5b = 40 \\[1em] \Rightarrow b = -8.\end{aligned}$$
Substituting value of b = -8 in (i) we get,
c = -5 - b = -5 - (-8) = 3.
Substituting b = -8 and c = 3 in 4x3 - bx2 + x - c we get,
Expression = 4x3 + 8x2 + x - 3
Dividing, 4x3 + 8x2 + x - 3 by (x + 1),
$$\begin{array}{l} \phantom{x + 1)}{4x^2 + 4x - 3} \\ x + 1\overline{\smash{\big)}4x^3 + 8x^2 + x - 3} \\ \phantom{x + 1}\underline{\underset{-}{}4x^3 \underset{-}{+} 4x^2} \\ \phantom{{x + 1}3x^3+}4x^2 + x \\ \phantom{{x + 1}3x^3\enspace\space}\underline{\underset{-}{}4x^2 \underset{-}{+} 4x} \\ \phantom{{x + 1}{3x^3+1}{+2}}-3x - 3 \\ \phantom{{x + 1}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}3x \underset{+}{-} 3} \\ \phantom{{x + 1}{2x^3+}{+2x^2-}{4x}}\times \end{array}$$
we get, quotient = 4x2 + 4x - 3.
Factorising 4x2 + 4x - 3,
= 4x2 + 6x - 2x - 3
= 2x(2x + 3) - 1(2x + 3)
= (2x - 1)(2x + 3).
Hence, b = -8, c = 3 and x3 + 8x2 + x - 3 = (x + 1)(2x - 1)(2x + 3).
If x + a is a common factor of expressions f(x) = x2 + px + q and g(x) = x2 + mx + n; show that : a = $\frac{n - q}{m - p}$
x + a = 0 ⇒ x = -a.
Since, (x + a) is factor of f(x) and g(x).
∴ f(-a) = g(-a)
⇒ (-a)2 + p(-a) + q = (-a)2 + m(-a) + n
⇒ a2 - pa + q = a2 - ma + n
⇒ a2 - a2 - pa + ma = n - q
⇒ ma - pa = n - q
⇒ a(m - p) = n - q
⇒ a = $\frac{n - q}{m - p}$.
Hence, proved that a = $\frac{n - q}{m - p}$.
The remainder, when x3 - x2 + x - 1 is divided by x + 1, is :
By remainder theorem,
The remainder theorem states that when a polynomial f(x) is divided by (x - a), then the remainder = f(a).
Given,
⇒ x + 1 = 0
⇒ x = -1.
Substituting x = -1 in x3 - x2 + x - 1 , we get :
⇒ (-1)3 - (-1)2 + (-1) - 1
⇒ -1 - 1 - 1 - 1
⇒ -4.
Hence, Option 2 is the correct option.
The number of common factor(s) of two polynomials x2 - 9 and x3 - x2 - 9x + 9 is /are :
Factorizing x2 - 9,
⇒ (x + 3)(x - 3)
Factorizing x3 - x2 - 9x + 9,
⇒ x2(x - 1) - 9(x - 1)
⇒ (x2 - 9)(x - 1)
⇒ (x + 3)(x - 3)(x - 1)
The number of common factor(s) of two polynomials x2 - 9 and x3 - x2 - 9x + 9 are two i.e.,(x + 3) and (x - 3).
Hence, Option 2 is the correct option.
Is (x - 2) a factor of x3 - 4x2 - 11x + 30 ?
Given,
⇒ x - 2 = 0
⇒ x = 2.
Substituting x = 2 in x3 - 4x2 - 11x + 30, we get :
⇒ 23 - 4(2)2 - 11(2) + 30
⇒ 8 - 4(4) - 22 + 30
⇒ 8 - 16 - 22 + 30
⇒ 38 - 38
⇒ 0.
Since, remainder = 0.
∴ x - 2 is factor of x3 - 4x2 - 11x + 30.
Hence, Option 1 is the correct option.
4x2 - kx + 5 leaves a remainder 2 when divided by x - 1. The value of k is :
By remainder theorem,
If polynomial ƒ(x) is divided by (x - a) then the remainder will be ƒ(a).
Given,
4x2 - kx + 5 leaves a remainder 2 when divided by x - 1.
∴ 4(1)2 - k(1) + 5 = 2
⇒ 4(1) - k + 5 = 2
⇒ 4 - k + 5 = 2
⇒ 9 - k = 2
⇒ k = 9 - 2 = 7.
Hence, Option 3 is the correct option.
If mx2 - nx + 8 has x - 2 as a factor, then :
By factor theorem,
If x - a is the factor of polynomial f(x), then remainder f(a) = 0.
Given,
x - 2 is a factor of mx2 - nx + 8.
Then substituting x = 2 in polynomial we get, remainder = 0 :
⇒ m(2)2 - 2n + 8 = 0
⇒ 4m - 2n + 8 = 0
⇒ 2(2m - n + 4) = 0
⇒ 2m - n + 4 = 0
⇒ n - 2m = 4.
Hence, Option 4 is the correct option.
Two polynomials x36 - 3x35 and x - 3.
Assertion (A) : If x - 3 is a factor of x36 - 3x35, the remainder is zero.
Reason (R) : The polynomial x - a is factor of polynomial p(x) = x36 - ax35, if p(a) = 0
options
Both A and R are true and R is correct reason for A.
Reason
By factor theorem,
(x - a) is a factor of the polynomial f(x), if the remainder i.e. f(a) = 0.
Let f(x) = x36 - 3x35
When, x - 3 is a factor of f(x), then f(3) = 0, by factor theorem.
∴ Assertion (A) is true.
When, p(x) = x36 - ax35 is divided by x - a, we get :
Remainder, p(a) = a36 - a.a35
= a36 - a36
= 0.
Since, p(a) = 0, thus (x - a) is factor of p(x).
∴ Reason (R) is true.
Thus, Both A and R are true and R is correct reason for R.
Hence, option 3 is the correct option.
The polynomial 3x3 + 8x2 - 15x + k and one of its factors as (x - 1).
Assertion (A) : The value of k = 4.
Reason (R) : x - 1 = 0 ⇒ x = 1.
∴ 3.(1)3 + 8.(1)2 - 15 x (1) + k = 0
options
Both A and R are true and R is correct reason for A.
Reason
Let, f(x) = 3x3 + 8x2 - 15x + k
By factor theorem,
(x - a) is a factor of the polynomial f(x), if the remainder i.e. f(a) = 0.
⇒ x - 1 = 0
⇒ x = 1.
Given,
x - 1 is one of the factors of f(x).
∴ f(1) = 0
⇒ 3.(1)3 + 8.(1)2 - 15.1 + k = 0
⇒ 3.1 + 8.1 - 15 + k = 0
⇒ 3 + 8 - 15 + k = 0
⇒ -4 + k = 0
⇒ k = 4.
Thus, both A and R are true and R is correct reason for A.
Hence, Option 3 is the correct option.
The polynomial x2 + x + b has (x + 3) as a factor of it.
Statement 1: The value of b is -4.
Statement 2: (x + 3) is a factor of x2 + x + b ⇒ (3)2 + 3 + b = 0.
option
Both the statements are false.
Reason
By factor theorem,
(x - a) is a factor of the polynomial f(x), if the remainder i.e. f(a) = 0.
⇒ x + 3 = 0
⇒ x = -3
Let, f(x) = x2 + x + b
Given,
The polynomial x2 + x + b has (x + 3) as a factor of it.
⇒ (-3)2 + (-3) + b = 0
⇒ 9 - 3 + b = 0
⇒ 6 + b = 0
⇒ b = -6.
∴ Statement 1 is incorrect.
Since,
⇒ (-3)2 + (-3) + b = 0
∴ Statement 2 is incorrect.
Hence, option 2 is the correct option.
When x3 + 3x2 - mx + 4 is divided by x - 2, the remainder is m + 3. Find the value of m.
Given,
When x3 + 3x2 - mx + 4 is divided by x - 2, the remainder is m + 3.
∴ (2)3 + 3(2)2 - m(2) + 4 = m + 3
⇒ 8 + 12 - 2m + 4 = m + 3
⇒ 24 - 2m = m + 3
⇒ m + 2m = 24 - 3
⇒ 3m = 21
⇒ m = 7.
Hence, m = 7.
What should be subtracted from 3x3 - 8x2 + 4x - 3, so that the resulting expression has x + 2 as a factor ?
The number to be subtracted = Remainder obtained on dividing 3x3 - 8x2 + 4x - 3 by x + 2.
x + 2 = 0 ⇒ x = -2.
Substituting x = -2 in 3x3 - 8x2 + 4x - 3 we get,
= 3(-2)3 - 8(-2)2 + 4(-2) - 3
= 3(-8) - 8(4) - 8 - 3
= -24 - 32 - 8 - 3
= -67.
Hence, no. to be subtracted = -67.
If (x + 1) and (x - 2) are factors of x3 + (a + 1)x2 - (b - 2)x - 6, find the values of a and b. And then, factorise the given expression completely.
Given,
(x + 1) and (x - 2) are factors of x3 + (a + 1)x2 - (b - 2)x - 6
x + 1 = 0 ⇒ x = -1.
Since, x + 1 is factor of x3 + (a + 1)x2 - (b - 2)x - 6. Hence, on substituting x = -1 in expression, remainder = 0.
⇒ (-1)3 + (a + 1)(-1)2 - (b - 2)(-1) - 6 = 0
-1 + (a + 1)(1) - (-b + 2) - 6 = 0
-1 + a + 1 + b - 2 - 6 = 0
a + b = 8
b = 8 - a ........(i)
x - 2 = 0 ⇒ x = 2.
Since, x - 2 is factor of x3 + (a + 1)x2 - (b - 2)x - 6. Hence, on substituting x = 2 in expression, remainder = 0.
⇒ (2)3 + (a + 1)(2)2 - (b - 2)(2) - 6 = 0
⇒ 8 + (a + 1)(4) - (2b - 4) - 6 = 0
⇒ 8 + 4a + 4 - 2b + 4 - 6 = 0
⇒ 10 + 4a - 2b = 0
⇒ 2b = 4a + 10
⇒ b = 2a + 5 ........(ii)
From (i) and (ii) we get,
⇒ 8 - a = 2a + 5
⇒ 2a + a = 8 - 5
⇒ 3a = 3.
⇒ a = 1.
Substituting a = 1, in (i) we get,
⇒ b = 8 - 1 = 7.
Substituting a = 1, b = 7 in expression we get,
Expression = x3 + (1 + 1)x2 - (7 - 2)x - 6 = x3 + 2x2 - 5x - 6.
On dividing, x3 + 2x2 - 5x - 6 by (x + 1),
$$\begin{array}{l} \phantom{x + 1)}{x^2 + x - 6} \\ x + 1\overline{\smash{\big)}x^3 + 2x^2 - 5x - 6} \\ \phantom{x + 1}\underline{\underset{-}{}x^3 \underset{+}{-} x^2} \\ \phantom{{x + 1}3x^3+}x^2 - 5x \\ \phantom{{x + 1}3x^3\enspace}\underline{\underset{-}{}x^2 \underset{-}{+} x} \\ \phantom{{x + 1}{x^32x^2}{+2}}-6x - 6 \\ \phantom{{x + 1}{x^32x^2}{+2x}}\underline{\underset{+}{-}6x \underset{+}{-} 6} \\ \phantom{{x + 1}{x^32x^2}{+2x^2-}}\times \end{array}$$
we get, quotient = x2 + x - 6.
Factorising, x2 + x - 6
= x2 + 3x - 2x - 6
= x(x + 3) - 2(x + 3)
= (x - 2)(x + 3).
Hence, a = 1, b = 7 and x3 + 2x2 - 5x - 6 = (x + 1)(x - 2)(x + 3).
If x - 2 is a factor of x2 + ax + b and a + b = 1, find the values of a and b.
Given,
x - 2 is a factor of x2 + ax + b.
Hence, on substituting x = 2 in x2 + ax + b, remainder = 0.
⇒ (2)2 + a(2) + b = 0
⇒ 4 + 2a + b = 0
⇒ b = -(2a + 4)
Substituting value of b in a + b = 1 we get,
⇒ a + -(2a + 4) = 1
⇒ a - 2a - 4 = 1
⇒ -a - 4 = 1
⇒ -a = 1 + 4
⇒ a = -5.
⇒ b = -(2a + 4) = -[2(-5) + 4] = -(-10 + 4) = 6.
Hence, a = -5 and b = 6.
Find the value of 'm', if mx3 + 2x2 - 3 and x2 - mx + 4 leave the same remainder when divided by x - 2.
x - 2 = 0 ⇒ x = 2.
Given,
mx3 + 2x2 - 3 and x2 - mx + 4 leave the same remainder when divided by x - 2.
∴ m(2)3 + 2(2)2 - 3 = (2)2 - m(2) + 4
8m + 8 - 3 = 4 - 2m + 4
8m + 2m = 8 - 8 + 3
10m = 3
m = $\frac{3}{10}$.
Hence, m = $\frac{3}{10}$.
The polynomial px3 + 4x2 - 3x + q is completely divisible by x2 - 1; find the values of p and q. Also for these values of p and q factorize the given polynomial completely.
x2 - 1 is a factor of px3 + 4x2 - 3x + q.
∴ (x - 1) and (x + 1) are factors of px3 + 4x2 - 3x + q.
Hence, substituting x = 1, -1 remainder = 0..
⇒ p(1)3 + 4(1)2 - 3(1) + q = 0
⇒ p + 4 - 3 + q = 0
⇒ p + q = -1
⇒ p = -1 - q .......(i)
p(-1)3 + 4(-1)2 - 3(-1) + q = 0
⇒ -p + 4 + 3 + q = 0
⇒ p = 7 + q .......(ii)
From (i) and (ii) we get,
⇒ -1 - q = 7 + q
⇒ 2q = -1 - 7
⇒ 2q = -8
⇒ q = -4.
Substituting q = -4 in (i) we get,
⇒ p = -1 - (-4) = -1 + 4 = 3.
Substituting p = 3 and q = -4 in px3 + 4x2 - 3x + q,
= 3x3 + 4x2 - 3x - 4.
On dividing, 3x3 + 4x2 - 3x - 4 by x - 1,
$$\begin{array}{l} \phantom{x - 1)}{3x^2 + 7x + 4} \\ x - 1\overline{\smash{\big)}3x^3 + 4x^2 - 3x - 4} \\ \phantom{x - 1}\underline{\underset{-}{}3x^3 \underset{+}{-} 3x^2} \\ \phantom{{x - 1}3x^3+3}7x^2 - 3x \\ \phantom{{x - 1}3x^3+}\underline{\underset{-}{}7x^2 \underset{+}{-} 7x} \\ \phantom{{x - 1}{3x^3+3x^2+}{+2}}4x - 4 \\ \phantom{{x - 1}{3x^3+3x^2+}\enspace \space}\underline{\underset{-}{}4x \underset{+}{-} 4} \\ \phantom{{x - 1}{3x^3+3x^2+}{+2-}}\times \end{array}$$
we get, quotient = 3x2 + 7x + 4
Factorising 3x2 + 7x + 4,
= 3x2 + 3x + 4x + 4
= 3x(x + 1) + 4(x + 1)
= (3x + 4)(x + 1).
Hence, p = 3, q = -4 and 3x3 + 4x2 - 3x - 4 = (x - 1)(x + 1)(3x + 4).
When the polynomial x3 + 2x2 - 5ax - 7 is divided by (x - 1), the remainder is A and when the polynomial x3 + ax2 - 12x + 16 is divided by (x + 2), the remainder is B. Find the value of 'a' if 2A + B = 0.
Given,
When the polynomial x3 + 2x2 - 5ax - 7 is divided by (x - 1), remainder is A.
x - 1 = 0 ⇒ x = 1
Substituting x = 1 in x3 + 2x2 - 5ax - 7 will give, remainder = A.
∴ (1)3 + 2(1)2 - 5a(1) - 7 = A
⇒ 1 + 2 - 5a - 7 = A
⇒ A = -(4 + 5a) ......(i)
Given,
When the polynomial x3 + ax2 - 12x + 16 is divided by (x + 2), remainder is B.
∴ x3 + ax2 - 12x + 16 = B
⇒ (-2)3 + a(-2)2 - 12(-2) + 16 = B
⇒ -8 + 4a + 24 + 16 = B
⇒ B = 32 + 4a ......(ii)
Given, 2A + B = 0
∴ -2(4 + 5a) + 32 + 4a = 0
⇒ -8 - 10a + 32 + 4a = 0
⇒ 24 - 6a = 0
⇒ 6a = 24
⇒ a = 4.
Hence, a = 4.
(3x + 5) is a factor of the polynomial (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15. Find the value of 'a'. For this value of 'a', factorise the given polynomial completely.
3x + 5 = 0 ⇒ x = -$\frac{5}{3}$
Since, (3x + 5) is a factor of the polynomial (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15. Substituting x = -$\frac{5}{3}$ in (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15, remainder = 0.
$$\begin{aligned}\Rightarrow (a - 1)\Big(-\frac{5}{3}\Big)^3 + (a + 1)\Big(-\frac{5}{3}\Big)^2 - (2a + 1)\Big(-\frac{5}{3}\Big) - 15 = 0 \\[1em] \Rightarrow (a - 1)\Big(-\frac{125}{27}\Big) + (a + 1)\Big(\frac{25}{9}\Big) + \frac{10a + 5}{3} - 15 = 0 \\[1em] \Rightarrow \frac{-125a + 125}{27} + \frac{25a + 25}{9} + \frac{10a + 5}{3} = 15 \\[1em] \Rightarrow \frac{-125a + 125 + 75a + 75 + 90a + 45}{27} = 15 \\[1em] \Rightarrow 40a + 245 = 405 \\[1em] \Rightarrow 40a = 160 \\[1em] \Rightarrow a = 4.\end{aligned}$$
Substituting a = 4 in (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15,
⇒ (4 - 1)x3 + (4 + 1)x2 - (2(4) + 1)x - 15
⇒ 3x3 + 5x2 - 9x - 15
⇒ x2(3x + 5) - 3(3x + 5)
⇒ (x2 - 3)(3x + 5)
⇒ $(x - \sqrt{3})(x + \sqrt{3})(3x + 5)$
Hence, a = 4 and 3x3 + 5x2 - 9x - 15 = $(x - \sqrt{3})(x + \sqrt{3})(3x + 5)$.
Using remainder theorem, find the value of k if on dividing 2x3 + 3x2 - kx + 5 by x - 2, leaves a remainder 7.
x - 2 = 0 ⇒ x = 2.
Given,
On dividing 2x3 + 3x2 - kx + 5 by x - 2, remainder = 7.
∴ 2(2)3 + 3(2)2 - k(2) + 5 = 7
⇒ 2(8) + 3(4) - 2k + 5 = 7
⇒ 16 + 12 - 2k + 5 = 7
⇒ 33 - 2k = 7
⇒ 2k = 33 - 7
⇒ 2k = 26
⇒ k = 13.
Hence, k = 13.
Find the value of 'a' if (x - a) is a factor of polynomial 3x3 + x2 - ax - 81.
By factor theorem,
If x - a is the factor of polynomial f(x), then remainder f(a) = 0.
Given,
f(x) = 3x3 + x2 - ax - 81
Then substituting x = a in polynomial we get, remainder = 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, a = 3.
While factorizing a given polynomial, using remainder and factor theorem, a student finds that x + 3 is a factor of 2x3 - x2 - 5x - 2.
(a) Is the student's, solution correct stating that (x + 3) is a factor of the given polynomial?
(b) Give a valid reason for your answer.
(c) Factorize the given polynomial completely.
Given polynomial: 2x3 - x2 - 5x - 2.
By factor theorem,
If x - a is the factor of polynomial f(x), then remainder f(a) = 0.
If (x + 3) is a factor, then by Factor Theorem :
f(−3) = 0.
Substituting x = -3 in polynomial we get, remainder = 0 :
⇒ 2(-3)3 - (-3)2 - 5(-3) - 2
⇒ 2(-27) - 9 + 15 - 2
⇒ -54 - 9 + 15 - 2
⇒ -50.
Since f(-3) ≠ 0, Remainder ≠ 0.
Hence, the student's solution is incorrect.
Factorizing,
Substituting x = 2 in polynomial we get,
f(2) = 2(2)3 - (2)2 - 5(2) - 2
= 16 - 4 - 10 - 2
= 16 - 16
= 0.
Since f(2) = 0, (x - 2) is a factor 2x3 - x2 - 5x - 2.
On dividing, 2x3 - x2 - 5x - 2 by x - 2,
$$\begin{array}{l} \phantom{x - 2)}{2x^2 + 3x + 1} \\ x - 2\overline{\smash{\big)}2x^3 - x^2 - 5x - 2} \\ \phantom{x - 1}\underline{\underset{-}{}2x^3 \underset{+}{-} 4x^2} \\ \phantom{{x - 1}3xm.^30}3x^2 - 5x \\ \phantom{{x - 1}3.3+}\underline{\underset{-}{}3x^2 \underset{+}{-} 6x} \\ \phantom{{x - 1}{3x^3+3x^2+}{+2}}x - 2 \\ \phantom{{x - 1}{3x^3+3x^2+}\enspace \space}\underline{\underset{-}{}x \underset{+}{-} 2} \\ \phantom{{x - 1}{3x^3+3x^2+}{+2-}}\times \end{array}$$
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)(x + 1)(2x + 1).
Hence, 2x3 - x2 - 5x - 2 = (x - 2)(x + 1)(2x + 1).
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