Factorization of Polynomial — Exercise 8(B)

For the polynomial x⁵ - x⁴ + x³ - 8x² + 6x + 15, the maximum number of linear factors is :

1. 9

2. 6

3. 7

4. 5

Answer

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 :

1. 8

2. 16

3. -8

4. -16

Answer

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 :

1. 49

2. 1

3. 0

4. -1

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 x²⁵ + x²⁴, we get :

⇒ (-1)²⁵ + (-1)²⁴

⇒ -1 + 1

⇒ 0.

Hence, Option 3 is the correct option.

Factors of 3x³ - 2x² - 8x are :

1. x(3x² - 2x - 8)

2. x(x - 2)(3x + 4)

3. 2x(x - 4)(2x + 1)

4. x(x - 4)(2x + 1)

Answer

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 :

1. (2 + x)(2 - x)(1 + x)

2. (x - 2)(1 + x)(2 + x)

3. (x + 2)(x - 2)(1 - x)

4. (2 + x)(x - 1)(2 - x)

Answer

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.

Answer

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.

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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.

Answer

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.

Answer

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 = -$\dfrac{3}{2}$,

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, $-\dfrac{3}{2}$.

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).

Answer

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.

Answer

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 = $\dfrac{3}{2}$ in 4x³ - bx² + x - c, value = 30.

$\Rightarrow 4\Big(\dfrac{3}{2}\Big)^3 - b\Big(\dfrac{3}{2}\Big)^2 + \Big(\dfrac{3}{2}\Big) - c = 30 \\[1em] \Rightarrow 4\Big(\dfrac{27}{8}\Big) - b\Big(\dfrac{9}{4}\Big) + \Big(\dfrac{3}{2}\Big) - c = 30 \\[1em] \Rightarrow \dfrac{27}{2} - \dfrac{9b}{4} + \dfrac{3}{2} - 30 = c \\[1em] \Rightarrow \dfrac{54 - 9b + 6 - 120}{4} = c \\[1em] \Rightarrow \dfrac{-9b - 60}{4} = c ........(ii)$

From (i) and (ii) we get,

$\Rightarrow -5 - b = \dfrac{-9b - 60}{4} \\[1em] \Rightarrow -20 - 4b = -9b - 60 \\[1em] \Rightarrow -20 + 60 = -9b + 4b \\[1em] \Rightarrow -5b = 40 \\[1em] \Rightarrow b = -8.$

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 = $\dfrac{n - q}{m - p}$

Answer

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 = $\dfrac{n - q}{m - p}$.

Hence, proved that a = $\dfrac{n - q}{m - p}$.