Factorization of Polynomial — Test Yourself

The remainder, when x³ - x² + x - 1 is divided by x + 1, is :

1. 0

2. -4

3. 2

4. 4

Answer

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 :

1. one

2. two

3. three

Answer

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 ?

1. yes

2. no

3. nothing can be said

4. none of the above is true

Answer

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 :

1. -6

2. 6

3. 7

4. -7

Answer

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 :

1. 2m - n = 4

2. 2m + n = 4

3. 2n + m = 4

4. n - 2m = 4

Answer

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

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

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

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

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

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

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.

Answer

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 ?

Answer

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.

Answer

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.

Answer

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.

Answer

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 = $\dfrac{3}{10}$.

Hence, m = $\dfrac{3}{10}$.

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.

Answer

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.

Answer

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.

Answer

3x + 5 = 0 ⇒ x = -$\dfrac{5}{3}$

Since, (3x + 5) is a factor of the polynomial (a - 1)x³ + (a + 1)x² - (2a + 1)x - 15. Substituting x = -$\dfrac{5}{3}$ in (a - 1)x³ + (a + 1)x² - (2a + 1)x - 15, remainder = 0.

$\Rightarrow (a - 1)\Big(-\dfrac{5}{3}\Big)^3 + (a + 1)\Big(-\dfrac{5}{3}\Big)^2 - (2a + 1)\Big(-\dfrac{5}{3}\Big) - 15 = 0 \\[1em] \Rightarrow (a - 1)\Big(-\dfrac{125}{27}\Big) + (a + 1)\Big(\dfrac{25}{9}\Big) + \dfrac{10a + 5}{3} - 15 = 0 \\[1em] \Rightarrow \dfrac{-125a + 125}{27} + \dfrac{25a + 25}{9} + \dfrac{10a + 5}{3} = 15 \\[1em] \Rightarrow \dfrac{-125a + 125 + 75a + 75 + 90a + 45}{27} = 15 \\[1em] \Rightarrow 40a + 245 = 405 \\[1em] \Rightarrow 40a = 160 \\[1em] \Rightarrow a = 4.$

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.

Answer

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.

Answer

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 = $\dfrac{81}{3}$

⇒ 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.

Answer

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).