Factorisation — Exercise 7

Find the remainder (without division) on dividing f(x) by (x - 2) where

(i) f(x) = 5x² - 7x + 4

(ii) f(x) = 2x³ - 7x² + 3

Answer

(i) By remainder theorem, on dividing f(x) by (x - a) , remainder = f(a)

∴ On dividing, f(x) = 5x² - 7x + 4 by (x - 2)

Remainder = f(2)

$= 5(2)^2 - 7(2) + 4 \\[0.5em] = 5(4) - 14 + 4 \\[0.5em] = 20 - 10 \\[0.5em] = 10.$

Hence, the value of remainder is 10.

(ii) By remainder theorem, on dividing f(x) by (x - a) , remainder = f(a)

∴ On dividing, f(x) = 2x³ - 7x² + 3 by (x - 2)

Remainder = f(2)

$= 2(2)^3 - 7(2)^2 + 3 \\[0.5em] = 2(8) - 28 + 3 \\[0.5em] = 16 - 28 + 3 \\[0.5em] = -9.$

Hence, the value of remainder is -9.

Using remainder theorem, find the remainder on dividing f(x) by (x + 3) where

(i) f(x) = 2x² - 5x + 1

(ii) f(x) = 3x³ + 7x² - 5x + 1

Answer

(i) By remainder theorem, on dividing f(x) by (x - a) , remainder = f(a)

∴ On dividing, f(x) = 2x² - 5x + 1 by (x + 3) or (x - (-3))

Remainder = f(-3)

$= 2(-3)^2 - 5(-3) + 1 \\[0.5em] = 2(9) + 15 + 1 \\[0.5em] = 18 + 16 \\[0.5em] = 34.$

Hence, the value of remainder is 34.

(ii) By remainder theorem, on dividing f(x) by (x - a) , remainder = f(a)

∴ On dividing, f(x) = 3x³ + 7x² - 5x + 1 by (x + 3) or (x - (-3))

Remainder = f(-3)

$= 3(-3)^3 + 7(-3)^2 - 5(-3) + 1 \\[0.5em] = -81 + 63 + 15 + 1 \\[0.5em] = -2. \\[0.5em]$

Hence, the value of remainder is -2.

Find the remainder (without division) on dividing f(x) by (2x + 1) where

(i) f(x) = 4x² + 5x + 3

(ii) f(x) = 3x³ - 7x² + 4x + 11

Answer

(i) By remainder theorem, on dividing f(x) by (x - a) , remainder = f(a)

∴ On dividing, f(x) = 4x² + 5x + 3 by (2x + 1) or 2(x - (-$\dfrac{1}{2}$))

Remainder = f(-$\dfrac{1}{2}$)

$= 4\big(-\dfrac{1}{2}\big)^2 + 5\big(-\dfrac{1}{2}\big) + 3 \\[1em] = 4\big(\dfrac{1}{4}\big) -\dfrac{5}{2} + 3 \\[1em] = 1 - \dfrac{5}{2} + 3 \text{ (On taking L.C.M.)} \\[1em] = \dfrac{2 - 5 + 6}{2} \\[1em] = \dfrac{3}{2} \\[1em] = 1\dfrac{1}{2}.$

Hence, the value of remainder is 1 $\dfrac{1}{2}$.

(ii) By remainder theorem, on dividing f(x) by (x - a) , remainder = f(a)

∴ On dividing, f(x) = 3x³ - 7x² + 4x + 11 by (2x + 1) or 2(x - (-$\dfrac{1}{2}$))

Remainder = f$\big(-\dfrac{1}{2}\big)$

$= 3\big(-\dfrac{1}{2}\big)^3 - 7\big(-\dfrac{1}{2}\big)^2 + 4\big(-\dfrac{1}{2}\big) + 11 \\[1em] = 3\big(-\dfrac{1}{8}\big) - 7\big(\dfrac{1}{4}\big) - 2 + 11 \\[1em] = -\dfrac{3}{8} - \dfrac{7}{4} + 9 \\[1em] = \dfrac{-3 - 14 + 72}{8} \\[1em] = \dfrac{55}{8} \\[1em] = 6\dfrac{7}{8}$

Hence, the value of remainder is $6\dfrac{7}{8}$.

Using remainder theorem, find the value of k if on dividing 2x³ + 3x² - kx + 5 by (x - 2) leaves a remainder 7.

Answer

By remainder theorem, on dividing f(x) by (x - a) , remainder = f(a)

∴ On dividing, f(x) = 2x³ + 3x² - kx + 5 by (x - 2)

Remainder = f(2)

Given, remainder = 7

$\therefore 2(2)^3 + 3(2)^2 - k(2) + 5 = 7 \\[0.5em] \Rightarrow 2(8) + 3(4) - 2k + 5 = 7 \\[0.5em] \Rightarrow 16 + 12 + 5 - 2k = 7 \\[0.5em] \Rightarrow 33 - 2k = 7 \\[0.5em] \Rightarrow 2k = 33 - 7 \\[0.5em] \Rightarrow 2k = 26 \\[0.5em] k = 13$

Hence, the value of k is 13.

Using remainder theorem, find the value of a if the division of x³ + 5x² - ax + 6 by (x - 1) leaves the remainder 2a.

Answer

By remainder theorem, on dividing f(x) by (x - a) , remainder = f(a)

∴ On dividing, f(x) = x³ + 5x² - ax + 6 by (x - 1)

Remainder = f(1)

Given, remainder = 2a

$\therefore (1)^3 + 5(1)^2 - a(1) + 6 = 2a \\[0.5em] \Rightarrow 1 + 5 - a + 6 = 2a \\[0.5em] \Rightarrow 12 - a = 2a \\[0.5em] \Rightarrow 3a = 12 \\[0.5em] a = 4 \\[0.5em]$

Hence, the value of a is 4.

What number must be subtracted from 2x² - 5x so that resulting polynomial leaves remainder 2 when divided by 2x + 1 ?

Answer

Let the number to be subtracted be a.

So, polynomial = 2x² - 5x - a

By remainder theorem, on dividing f(x) by (x - b) , remainder = f(b)

∴ On dividing, f(x) = 2x² - 5x - a by (2x + 1) or 2(x - $\big(-\dfrac{1}{2}\big)$)

Remainder = f$\big(-\dfrac{1}{2}\big)$

Given, remainder = 2

$\therefore 2\big(-\dfrac{1}{2}\big)^2 - 5\big(-\dfrac{1}{2}\big) - a = 2 \\[1em] \Rightarrow 2\big(\dfrac{1}{4}\big) + \dfrac{5}{2} - a = 2 \\[1em] \Rightarrow \dfrac{1}{2} + \dfrac{5}{2} - a = 2 \\[1em] \Rightarrow \dfrac{6}{2} - a = 2 \\[1em] \Rightarrow 3 - a = 2 \\[1em] \Rightarrow a = 3 - 2 \\[1em] a = 1.$

Hence, the value of a is 1.

What number must be added to 2x³ - 3x² - 8x so that resulting polynomial leaves the remainder 10 when divided by 2x + 1?

Answer

Given,

⇒ 2x + 1 = 0

⇒ 2x = -1

⇒ x = $-\dfrac{1}{2}$

Let number added be a.

Polynomial = 2x³ - 3x² - 8x + a

By remainder theorem,

When a polynomial p(x) is divided by (x - a), then the remainder = f(a).

$\therefore 2.\Big(-\dfrac{1}{2}\Big)^3 - 3.\Big(-\dfrac{1}{2}\Big)^2 - 8.\Big(-\dfrac{1}{2}\Big) + a = 10\\[1em] \Rightarrow 2.\Big(-\dfrac{1}{8}\Big) - 3.\Big(\dfrac{1}{4}\Big) + \Big(\dfrac{8}{2}\Big) + a = 10\\[1em] \Rightarrow -\dfrac{2}{8} - \Big(\dfrac{3}{4}\Big) + 4 + a = 10\\[1em] \Rightarrow -\dfrac{1}{4} - \Big(\dfrac{3}{4}\Big) + 4 + a = 10\\[1em] \Rightarrow \dfrac{-1 - 3}{4} + 4 + a = 10\\[1em] \Rightarrow \dfrac{-4}{4} + 4 + a = 10\\[1em] \Rightarrow -1 + 4 + a = 10\\[1em] \Rightarrow 3 + a = 10\\[1em] \Rightarrow a = 10 - 3\\[1em] \Rightarrow a = 7$

Hence, the number to be added = 7.

When divided by x - 3 the polynomials x³ - px² + x + 6 and 2x³ - x² - (p + 3)x - 6 leave the same remainder . Find the value of 'p'.

Answer

By remainder theorem, on dividing f(x) by (x - b), remainder = f(b)

∴ On dividing, f(x) = x³ - px² + x + 6 by (x - 3)

Remainder = 3³ - p(3²) + 3 + 6 = 27 - 9p + 9 = 36 - 9p.

∴ On dividing, f(x) = 2x³ - x² - (p + 3)x - 6 by (x - 3)

Remainder = 2(3)³ - 3² - (p + 3)(3) - 6 = 2(27) - 9 - 3p - 9 - 6 = 54 - 9 - 3p - 9 - 6 = 30 - 3p

According to question,

$\Rightarrow 36 - 9p = 30 - 3p \\[0.5em] \Rightarrow 36 - 30 = 9p - 3p \\[0.5em] \Rightarrow 6 = 6p \\[0.5em] \Rightarrow p = 1$

Hence, the value of p is 1.

Find 'a' if the two polynomials ax³ + 3x² - 9 and 2x³ + 4x + a, leaves the same remainder when divided by x + 3.

Answer

By remainder theorem, on dividing f(x) by (x - b), remainder = f(b)

∴ On dividing, f(x) = ax³ + 3x² - 9 by (x + 3) or (x - (-3))

Remainder = f(-3) = a(-3)³ + 3(-3)² - 9 = -27a + 27 - 9 = 18 - 27a

∴ On dividing, f(x) = f(-3) = 2x³ + 4x + a by (x + 3) or (x - (-3))

Remainder = 2(-3)³ + 4(-3) + a = -54 - 12 + a = a - 66

According to question,

$\Rightarrow 18 - 27a = a - 66 \\[0.5em] \Rightarrow a + 27a = 66 + 18 \\[0.5em] \Rightarrow 28a = 84 \\[0.5em] \Rightarrow a = \dfrac{84}{28} \\[0.5em] \Rightarrow a = 3.$

Hence, the value of p is 3.

The polynomials ax³ + 3x² - 3 and 2x³ - 5x + a when divided by x - 4 leave the remainder r₁ and r₂ respectively. If 2r₁ = r₂, then find the value of a.

Answer

By remainder theorem, on dividing f(x) by (x - b), remainder = f(b)

∴ On dividing, f(x) = ax³ + 3x² - 3 by (x - 4)

Remainder = f(4) = a(4)³ + 3(4)² - 3 = 64a + 45

∴ On dividing, f(x) = 2x³ - 5x + a by (x - 4)

Remainder = f(4) = 2(4)³ - 5(4) + a = 128 - 20 + a = 108 + a

According to question,

r₁ = 64a + 45

r₂ = 108 + a

2r₁ = r₂

$\therefore 2(64a + 45) = 108 + a \\[0.5em] \Rightarrow 128a + 90 = 108 + a \\[0.5em] \Rightarrow 128a - a = 108 - 90 \\[0.5em] \Rightarrow 127a = 18 \\[0.5em] \Rightarrow a = \dfrac{18}{127}.$

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 two remainders is 1.

Answer

By remainder theorem, on dividing f(x) by (x - b), remainder = f(b)

∴ On dividing, f(x) = x³ + (kx + 8)x + k by x + 1 or (x - (-1))

Remainder = r₁ = f(-1) = -1³ + ((-1)k + 8)(-1) + k
= -1 + (8 - k)(-1) + k
= -1 - 8 + k + k
= 2k - 9

∴ On dividing, f(x) = x³ + (kx + 8)x + k by x - 2

Remainder = r₂ = (2)³ + (k(2) + 8)2 + k
= 8 + 4k + 16 + k
= 5k + 24

Given, sum of two remainders = 1

∴ r₁ + r₂ = 1

$\Rightarrow 2k - 9 + 5k + 24 = 1 \\[0.5em] \Rightarrow 7k + 15 = 1 \\[0.5em] \Rightarrow 7k = 1 - 15 \\[0.5em] \Rightarrow 7k = -14 \\[0.5em] \Rightarrow k = -\dfrac{14}{7} \\[0.5em] k = -2.$

The first remainder is 2k - 9 and the second remainder is 5k + 24 and the value of k is -2.

By factor theorem, show that (x + 3) and (2x - 1) are the factors of 2x² + 5x - 3.

Answer

By factor theorem, (x - a) is a factor of f(x), if f(a) = 0.

f(x) = 2x² + 5x - 3

(x + 3) = (x - (-3)) is a factor of f(x), if f(-3) = 0

f(-3) = 2(-3)² + 5(-3) - 3
= 2(9) - 15 - 3
= 18 - 18 = 0

2x - 1 = 2(x - $\dfrac{1}{2}$) is a factor of f(x), if f($\dfrac{1}{2}$) = 0

$f\big(\dfrac{1}{2}\big) = 2\big(\dfrac{1}{2}\big)^2 + 5\big(\dfrac{1}{2}\big) - 3 \\[1em] = 2\big(\dfrac{1}{4}\big) + \dfrac{5}{2} - 3 \\[1em] = \dfrac{1}{2} + \dfrac{5}{2} - 3 \\[1em] = 3 - 3 = 0$

Since, f(-3) and f$\big(\dfrac{1}{2}\big)$ = 0 , hence, (x - 3) and (2x - 1) are factors of 2x² + 5x - 3.

Without actual division, prove that x⁴ + 2x³ - 2x² + 2x - 3 is exactly divisible by x² + 2x - 3.

Answer

Let, f(x) = x⁴ + 2x³ - 2x² + 2x - 3

g(x) = x² + 2x - 3
= x² + 3x - x - 3
= x(x + 3) - 1(x + 3)
= (x - 1)(x + 3)

$\Rightarrow$ (x - 1) and (x + 3) are factors of g(x).

In order to prove that f(x) is exactly divisible by g(x), it is sufficient to prove that x - 1 and x + 3 are factors of f(x) i.e. it is sufficient to show that f(1) = 0 and f(-3) = 0.

Now,
f(1) = (1)⁴ + 2(1)³ - 2(1)² + 2(1) - 3

= 1 + 2 - 2 + 2 - 3 = 0

f(-3) = (-3)⁴ + 2(-3)³ - 2(-3)² + 2(-3) - 3

= 81 - 54 - 18 - 6 - 3 = 0

∴ f(x) is divisible by (x - 1) and (x + 3)

Hence, f(x) is exactly divisible by g(x).

Show that (x - 2) is a factor 3x² - x - 10. Hence, factorize 3x² - x - 10.

Answer

By factor theorem, (x - a) is a factor of f(x), if f(a) = 0.

f(x) = 3x² - x - 10

(x - 2) is a factor of f(x), if f(2) = 0

f(2) = 3(2)² - 2 - 10

= 12 - 12 = 0

Hence, x - 2 is a factor of 3x² - x - 10.

Now, factorizing $3x^2 - x - 10$,

$\Rightarrow 3x^2 - 6x + 5x - 10 \\[0.5em] \Rightarrow 3x(x - 2) + 5(x - 2) \\[0.5em] \Rightarrow (3x + 5)(x - 2)$

Hence, 3x² - x - 10 = (x - 2)(3x + 5).

Using factor theorem, show that (x - 2) is a factor of x³ + x² - 4x - 4. Hence, factorise the polynomial completely.

Answer

By factor theorem, (x - a) is a factor of f(x), if f(a) = 0.

f(x) = x³ + x² - 4x - 4

(x - 2) is a factor of f(x), if f(2) = 0

$f(2) = (2)^3 + (2)^2 - 4(2) - 4 \\[0.5em] = 8 + 4 - 8 - 4 \\[0.5em] = 0$

Hence, (x - 2) is a factor of x³ + x² - 4x - 4.

Now, factorizing x³ + x² - 4x - 4,

$\Rightarrow x^2(x + 1) - 4(x + 1) \\[0.5em] \Rightarrow (x^2 - 4)(x + 1) \\[0.5em] \Rightarrow (x - 2)(x + 2)(x + 1)$

Hence, x³ + x² - 4x - 4 = (x - 2)(x + 1)(x + 2).

Show that 2x + 7 is a factor of 2x³ + 5x² - 11x - 14. Hence, factorise the given expression completely, using the factor theorem.

Answer

By factor theorem, (x - a) is a factor of f(x), if f(a) = 0.

f(x) = 2x³ + 5x² - 11x - 14

(2x + 7) or $2(x - \big(-\dfrac{7}{2}\big))$ is a factor of f(x), if $f\big(-\dfrac{7}{2}\big)$ = 0

$f\big(-\dfrac{7}{2}\big) = 2\big(-\dfrac{7}{2}\big)^3 + 5\big(-\dfrac{7}{2}\big)^2 - 11\big(-\dfrac{7}{2}\big) -14 \\[1em] = 2\big(-\dfrac{343}{8}\big) + 5\big(\dfrac{49}{4}\big) + \big(\dfrac{77}{2}\big) - 14 \\[1em] = \big(-\dfrac{343}{4}\big) + \big(\dfrac{245}{4}\big) + \big(\dfrac{77}{2}\big) - 14$

On taking LCM,

$= \big(\dfrac{-343 + 245 + 154 - 56}{4}\big) \\[1em] = \big(\dfrac{-399 + 399}{4}\big) \\[1em] = 0$

Hence, (2x + 7) is the factor of f(x).

On dividing f(x) by (2x + 7),

$\begin{array}{l} \phantom{2x + 7)}{x^2 - x - 2} \\ 2x + 7\overline{\smash{\big)}2x^3 + 5x^2 - 11x - 14} \\ \phantom{2x + 7}\underline{\underset{-}{}2x^3 \underset{-}{+} 7x^2} \\ \phantom{{2x + 7}2x^3+} -2x^2 - 11x \\ \phantom{{2x + 7}2x^3+}\underline{\underset{+}{-}2x^2 \underset{+}{-} 7x} \\ \phantom{{2x + 7}{2x^3+}{+2x^2-}}-4x - 14 \\ \phantom{{2x + 7}{2x^3+}{+2x^2-}}\underline{\underset{+}{-}4x \underset{+}{-} 14} \\ \phantom{{2x + 7}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, x² - x - 2 as the quotient and remainder = 0.

$\therefore 2x^3 + 5x^2 - 11x - 14 = (2x + 7)(x^2 - x - 2) \\[0.5em] = (2x + 7)(x^2 - 2x + x - 2) \\[0.5em] = (2x + 7)(x(x - 2) + 1(x - 2)) \\[0.5em] = (2x + 7)(x + 1)(x - 2)$

Hence, 2x³ + 5x² - 11x - 14 = (2x + 7)(x + 1)(x - 2)

Use factor theorem to factorise the following polynomials completely :

(i) x³ + 2x² - 5x - 6

(ii) x³ - 13x - 12

(iii) 6x³ + 17x² + 4x - 12

Answer

(i) f(x) = x³ + 2x² - 5x - 6

Putting, x = -1 in f(x)

$f(-1) = (-1)^3 + 2(-1)^2 - 5(-1) - 6 \\[0.5em] = -1 + 2 + 5 - 6 \\[0.5em] = 7 - 7 \\[0.5em] = 0$

Since, f(-1) = 0, hence (x + 1) is factor of f(x) by factor theorem.

Now, dividing f(x) by (x + 1),

we get x² + x - 6 as the quotient and remainder = 0.

$\therefore x^3 + 2x^2 - 5x - 6 = (x + 1)(x^2 + x - 6) \\[0.5em] = (x + 1)(x^2 + 3x - 2x - 6) \\[0.5em] = (x + 1)(x(x + 3) - 2(x + 3)) \\[0.5em] = (x + 1)(x - 2)(x + 3)$

Hence, x³ + 2x² - 5x - 6 = (x + 1)(x - 2)(x + 3).

(ii) Let f(x) = x³ - 13x - 12

Putting, x = 4 in f(x)

$f(4) = (4)^3 - 13(4) - 12 \\[0.5em] = 64 - 52 - 12 \\[0.5em] = 64 - 64 \\[0.5em] = 0$

Since, f(4) = 0, hence (x - 4) is factor of f(x) by factor theorem.

Now, dividing f(x) by (x - 4),

we get x² + 4x + 3 as quotient and remainder = 0.

$\therefore x^3 - 13x - 12 = (x - 4)(x^2 + 4x + 3) \\[0.5em] = (x - 4)(x^2 + 3x + x + 3) \\[0.5em] = (x - 4)(x(x + 3) + 1(x + 3)) \\[0.5em] = (x - 4)(x + 1)(x + 3)$

Hence, x³ - 13x - 12 = (x - 4)(x + 1)(x + 3).

(iii) Given,

f(x) = 6x³ + 17x² + 4x - 12

Substituting x = -2 in f(x), we get :

f(-2) = 6(-2)³ + 17(-2)² + 4(-2) - 12

= 6 × -8 + 17 × 4 - 8 - 12

= -48 + 68 - 8 - 12

= -68 + 68

= 0.

Since, f(-2) = 0, hence (x + 2) is factor of f(x).

Dividing f(x) by (x + 2), we get :

$\begin{array}{l} \phantom{x - 3x )}{\quad 6x^2 + 5x - 6} \\ x + 2\overline{\smash{\big)}\quad 6x^3 + 17x^2 + 4x - 12} \\ \phantom{x^2 + 4)}\phantom{)}\underline{\underset{-}{+}6x^3 \underset{-}{+} 12x^2 } \\ \phantom{{x^2 + 3x - 5 =d)}} 5x^2 + 4x - 12 \\ \phantom{{x^2 + 3x - 54 =)}}\underline{\underset{-}{+}5x^2 \underset{-}{+}10x } \\ \phantom{{x^2 + 3x - 54)} + 2x + 3 }-6x - 12\ \phantom{{x^2 + 3x - 54 =zccdvz)}}\underline {\underset{+}{-}6x \underset{+}{-}12 } \\ \phantom{{x^2 + 3x - 54)} + 2x + 3 + dc}\times\ \end{array}$

We get 6x² + 5x - 6 as the quotient and remainder = 0.

∴ 6x³ + 17x² + 4x - 12 = (x + 2)(6x² + 5x - 6)

= (x + 2)(6x² + 9x - 4x - 6)

= (x + 2)[3x(2x + 3) - 2(2x + 3)]

= (x + 2)(2x + 3)(3x - 2)

Hence,6x³ + 17x² + 4x - 12 = (x + 2)(2x + 3)(3x - 2).

Use Remainder Theorem to factorise the following polynomials completely :

(i) 2x³ + x² - 13x + 6

(ii) 3x³ + 2x² - 19x + 6

(iii) 2x³ + 3x² - 9x - 10

(iv) x³ + 10x² - 37x + 26

Answer

(i) Let f(x) = 2x³ + x² - 13x + 6

Putting, x = 2 in f(x)

$f(2) = 2(2)^3 + 2^2 - 13(2) + 6 \\[0.5em] = 16 + 4 - 26 + 6 \\[0.5em] = 0$

Since, f(2) = 0 , (x - 2) is factor of f(x) by factor theorem.
Dividing, f(x) 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, 2x² + 5x - 3 as quotient and remainder = 0.

$\therefore 2x^3 + x^2 - 13x + 6 = (x - 2)(2x^2 + 5x - 3) \\[0.5em] = (x - 2)(2x^2 + 6x - x - 3) \\[0.5em] = (x - 2)(2x(x + 3) - 1(x + 3)) \\[0.5em] = (x - 2)(2x - 1)(x + 3)$

Hence, 2x³ + x² - 13x + 6 = (x - 2)(2x - 1)(x + 3).

(ii) Let f(x) = 3x³ + 2x² - 19x + 6

Putting, x = 2 in f(x)

$f(2) = 3(2)^3 + 2(2)^2 - 19(2) + 6 \\[0.5em] = 24 + 8 - 38 + 6 \\[0.5em] = 38 - 38 \\[0.5em] = 0$

Since, f(2) = 0, (x - 2) is factor of f(x) by factor theorem.

Dividing, f(x) 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, 3x² + 8x - 3 as quotient and remainder = 0.

$\therefore 3x^3 + 2x^2 - 19x + 6 = (x - 2)(3x^2 + 8x - 3) \\[0.5em] = (x - 2)(3x^2 + 9x - x - 3) \\[0.5em] = (x - 2)(3x(x + 3) - 1(x + 3)) \\[0.5em] = (x - 2)(3x - 1)(x + 3)$

Hence, 3x³ + 2x² - 19x + 6 = (x - 2)(3x - 1)(x + 3).

(iii) Let f(x) = 2x³ + 3x² - 9x - 10

Putting, x = 2 in f(x)

$f(2) = 2(2)^3 + 3(2)^2 - 9(2) - 10 \\[0.5em] = 16 + 12 - 18 - 10 \\[0.5em] = 28 - 28 \\[0.5em] = 0$

Since, f(2) = 0, (x - 2) is factor of f(x) by factor theorem.

Dividing, f(x) by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{2x^2 + 7x + 5} \\ x - 2\overline{\smash{\big)}2x^3 + 3x^2 - 9x - 10} \\ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{-}{+}4x^2} \\ \phantom{{x - 2}2x^3+4}7x^2 - 9x \\ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}7x^2 \underset{+}{-} 14x} \\ \phantom{{x - 2}{2x^3+}{+2x^2}}5x - 10 \\ \phantom{{x - 2}{2x^3+}{+2x}}\underline{\underset{-}{ }5x \underset{+}{-} 10} \\ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, 2x² + 7x + 5 as quotient and remainder = 0.

$\therefore 2x^3 + 3x^2 - 9x - 10 = (x - 2)(2x^2 + 7x + 5) \\[0.5em] = (x - 2)(2x^2 + 5x + 2x + 5) \\[0.5em] = (x - 2)(x(2x + 5) + 1(2x + 5)) \\[0.5em] = (x - 2)(x + 1)(2x + 5)$

Hence, 2x³ + 3x² - 9x - 10 = (x - 2)(x + 1)(2x + 5).

(iv) Let f(x) = x³ + 10x² - 37x + 26

Putting, x = 1 in f(x)

$f(1) = (1)^3 + 10(1)^2 - 37(1) + 26 \\[0.5em] = 1 + 10 - 37 + 26 \\[0.5em] = 37 - 37 \\[0.5em] = 0$

Since, f(1) = 0 , (x - 1) is factor of f(x) by factor theorem.

Dividing, f(x) by (x - 1),

$\begin{array}{l} \phantom{x - 1)}{x^2 + 11x - 26} \\ x - 1\overline{\smash{\big)}x^3 + 10x^2 - 37x + 26} \\ \phantom{x - 1}\underline{\underset{-}{}x^3 \underset{+}{-}x^2} \\ \phantom{{x - 1}2x^3+4}11x^2 - 37x \\ \phantom{{x - 1}2x^3+}\underline{\underset{-}{}11x^2 \underset{+}{-} 11x} \\ \phantom{{x - 1}{2x^3+}{+11x^2}}-26x + 26 \\ \phantom{{x - 1}{2x^3+}{+11x^2}}\underline{\underset{+}{-}26x \underset{-}{+} 26} \\ \phantom{{x - 1}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, x² + 11x - 26 as quotient and remainder = 0.

$\therefore x^3 + 10x^2 - 37x + 26 = (x - 1)(x^2 + 11x - 26) \\[0.5em] = (x - 1)(x^2 + 13x - 2x - 26) \\[0.5em] = (x - 1)(x(x + 13) - 2(x + 13)) \\[0.5em] = (x - 1)(x - 2)(x + 13)$

Hence, x³ + 10x² - 37x + 26 = (x - 1)(x - 2)(x + 13).

If (2x + 1) is a factor of 6x³ + 5x² + ax - 2, find the value of a.

Answer

f(x) = 6x³ + 5x² + ax - 2

If, (2x + 1) or 2(x - (-$\dfrac{1}{2}$)) is a factor of f(x) then f(-$\dfrac{1}{2}$) = 0

$\therefore 6\big(-\dfrac{1}{2}\big)^3 + 5\big(-\dfrac{1}{2}\big)^2 + a(-\dfrac{1}{2}) - 2 = 0 \\[1em] \Rightarrow -\dfrac{3}{4} + \dfrac{5}{4} - \dfrac{a}{2} - 2 = 0$

On taking L.C.M.,

$\Rightarrow \dfrac{-3 + 5 - 2a - 8}{4} = 0 \\[0.5em] \Rightarrow \dfrac{-6 - 2a}{4} = 0 \\[0.5em]$

On Cross Multiplying,

$\Rightarrow -6 - 2a = 0 \\[0.5em] \Rightarrow 2a = -6 \\[0.5em] a = -3.$

Hence, the value of a is -3.

If (3x - 2) is a factor of 3x³ - kx² + 21x - 10, find the value of k.

Answer

f(x) = 3x³ - kx² + 21x - 10

If, (3x - 2) or $3(x - \big(\dfrac{2}{3}\big))$ is a factor of f(x) then f($\dfrac{2}{3}$) = 0

$\therefore 3\big(\dfrac{2}{3}\big)^3 - k\big(\dfrac{2}{3}\big)^2 + 21\big(\dfrac{2}{3}\big) - 10 = 0 \\[1em] \Rightarrow 3\big(\dfrac{8}{27}\big) - k\big(\dfrac{4}{9}\big) + 14 - 10 = 0 \\[1em] \Rightarrow \dfrac{8}{9} - \dfrac{4k}{9} + 4 = 0 \\[1em] \Rightarrow \dfrac{8 - 4k + 36}{9} = 0 \\[1em] \Rightarrow 44 - 4k = 0 \\[1em] \Rightarrow 4k = 44 \\[1em] k = 11.$

Hence, the value of k is 11.

If (x - 2) is a factor of 2x³ - x² - px - 2, then

(i) Find the value of p.

(ii) with this value of p, factorise the above expression completely.

Answer

(i) f(x) = 2x³ - x² - px - 2

If, (x - 2) is a factor of f(x), then f(2) = 0

$\therefore 2(2)^3 - (2)^2 - 2p - 2 = 0 \\[0.5em] \Rightarrow 16 - 4 - 2 - 2p = 0 \\[0.5em] \Rightarrow 10 - 2p = 0 \\[0.5em] \Rightarrow 2p = 10 \\[0.5em] p = 5.$

Hence, the value of p is 5.

(ii) Putting value of p = 5 in f(x),

f(x) = 2x³ - x² - 5x - 2

Since, (x - 2) is a factor of f(x), dividing f(x) by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{2x^2 + 3x + 1} \\ x - 2\overline{\smash{\big)}2x^3 - x^2 - 5x - 2} \\ \phantom{x - 2}\underline{\underset{-}{ }2x^3 \underset{+}{-} 4x^2} \\ \phantom{{x - 2}2x^3+4}3x^2 - 5x \\ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}3x^2 \underset{+}{-} 6x} \\ \phantom{{x - 2}{2x^3+}{+3x^2+5}}x - 2 \\ \phantom{{x - 2}{2x^3+}{+3x^2+5}}\underline{\underset{-}{ }x \underset{+}{-} 2} \\ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, 2x² + 3x + 1 as quotient and remainder = 0.

$\therefore 2x^3 - x^2 - 5x - 2 = (x - 2)(2x^2 + 3x + 1) \\[0.5em] = (x - 2)(2x^2 + 2x + x + 1) \\[0.5em] = (x - 2)(2x(x + 1) + 1(x + 1)) \\[0.5em] = (x - 2)(2x + 1)(x + 1)$

Hence, 2x³ - x² - 5x - 2 = (x - 2)(2x + 1)(x + 1).

What number should be subtracted from 2x³ - 5x² + 5x so that the resulting polynomial has 2x - 3 as a factor?

Answer

Let the number to be subtracted be a.

f(x) = 2x³ - 5x² + 5x - a

If, (2x - 3) or 2(x - ($\dfrac{3}{2}$)) is a factor of f(x) then f($\dfrac{3}{2}$) = 0, by factor theorem

$\therefore 2\big(\dfrac{3}{2}\big)^3 - 5\big(\dfrac{3}{2}\big)^2 + 5(\dfrac{3}{2}) - a = 0 \\[1em] \Rightarrow 2\big(\dfrac{27}{8}\big) - 5\big(\dfrac{9}{4}\big) + \dfrac{15}{2} - a = 0 \\[1em] \Rightarrow \dfrac{27}{4} - \dfrac{45}{4} + \dfrac{15}{2} - a = 0 \\[1em] \Rightarrow \dfrac{27 - 45 + 30 - 4a}{4} = 0 \\[1em] \Rightarrow 12 - 4a = 0 \\[1em] \Rightarrow 4a = 12 \\[1em] a = 3.$

Hence, the number to be subtracted is 3.

Find the value of the constants a and b, if (x - 2) and (x + 3) are both factors of the expression x³ + ax² + bx - 12.

Answer

f(x) = x³ + ax² + bx - 12

If (x - 2) and (x + 3) or (x - (-3)) are factors of f(x) then, f(2) and f(-3) = 0.

$\therefore f(2) = 2^3 + a(2)^2 + 2b - 12 = 0 \\[0.5em] \Rightarrow 8 + 4a + 2b - 12 = 0 \\[0.5em] \Rightarrow 4a + 2b - 4 = 0 \\[0.5em] \Rightarrow 4a + 2b = 4$

On dividing equation by 2,

$\Rightarrow 2a + b = 2 \\[0.5em] b = 2 - 2a \text{ \space (Equation 1)}$

$\therefore f(-3) = (-3)^3 + a(-3)^2 + (-3)b - 12 = 0 \\[0.5em] \Rightarrow -27 + 9a - 3b - 12 = 0 \\[0.5em] \Rightarrow 9a - 3b - 39 = 0$

Putting value of b = 2 - 2a from equation 1,

$\Rightarrow 9a - 3(2 - 2a) - 39 = 0 \\[0.5em] \Rightarrow 9a - 6 + 6a - 39 = 0 \\[0.5em] \Rightarrow 15a - 45 = 0 \\[0.5em] \Rightarrow 15a = 45 \\[0.5em] \Rightarrow a = 3 \\[0.5em] \therefore b = 2 - 2a = 2 - 6 = -4$

Hence, the value of a is 3 and that of b is -4.

If (x + 2) and (x + 3) are factors of x³ + ax + b, find the values of a and b.

Answer

f(x) = x³ + ax + b

If (x + 2) or (x - (-2)) and (x + 3) or (x - (-3)) are factors of f(x) then, f(-2) and f(-3) = 0.

$\therefore f(-2) = (-2)^3 + (-2)a + b = 0 \\[0.5em] \Rightarrow -8 - 2a + b = 0 \\[0.5em] \Rightarrow b = 2a + 8 \text{ \space (Equation 1)}$

$\therefore f(-3) = (-3)^3 + (-3)a + b = 0 \\[0.5em] \Rightarrow -27 - 3a + b = 0$

Putting value of b = 2a + 8 from equation 1,

$\Rightarrow -27 - 3a + 2a + 8 = 0 \\[0.5em] \Rightarrow -27 - a + 8 = 0 \\[0.5em] \Rightarrow -a - 19 = 0 \\[0.5em] \Rightarrow a = -19 \\[0.5em] \therefore b = 2a + 8 = -38 + 8 = -30$

Hence, the value of a is -19 and that of b is -30.

If (x + 2) and (x - 3) are the factors of x³ + ax + b, find the values of a and b. With these values of a and b, factorise the given expression.

Answer

f(x) = x³ + ax + b

If (x + 2) or (x - (-2)) and (x - 3) are factors of f(x) then, f(-2) and f(3) = 0.

$\therefore f(-2) = (-2)^3 + (-2)a + b = 0 \\[0.5em] \Rightarrow -8 - 2a + b = 0 \\[0.5em] \Rightarrow b = 2a + 8 \text{ \space (Equation 1)}$

$\therefore f(3) = (3)^3 + (3)a + b = 0 \\[0.5em] \Rightarrow 27 + 3a + b = 0$

Putting value of b = 2a + 8 from equation 1,

$\Rightarrow 27 + 3a + 2a + 8 = 0 \\[0.5em] \Rightarrow 35 + 5a = 0 \\[0.5em] \Rightarrow 5a = -35 \\[0.5em] \Rightarrow a = -7 \\[0.5em] \therefore b = 2a + 8 = -14 + 8 = -6$

Putting the values of a and b in f(x) we get,

f(x) = x³ - 7x - 6

Since, (x + 2) and (x - 3) are factors of f(x) hence, (x + 2)(x - 3) = (x² - x - 6) is also the factor.

On dividing f(x) by x² - x - 6,

$\begin{array}{l} \phantom{x^2 - x - 6)}{x + 1} \\ x^2 - x - 6\overline{\smash{\big)}x^3 - 7x - 6} \\ \phantom{x^2 - x - 6}\underline{\underset{-}{ }x^3 \underset{+}{-} x^2 \underset{+}{-} 6x} \\ \phantom{{x^2 - x - 6}2x^3+4}x^2 - x - 6 \\ \phantom{{x^2 - x - 6}2x^3+}\underline{\underset{-}{}x^2 \underset{+}{-} x \underset{+}{-} 6} \\ \phantom{{x^2 - x - 6}{2x^3+}{+2x^2-}}\times \end{array}$

we get, (x + 1) as quotient and remainder = 0.

$\therefore x^3 - 7x - 6 = (x^2 - x - 6)(x + 1) \\[0.5em] = (x^2- 3x + 2x - 6)(x + 1) \\[0.5em] = (x(x - 3) + 2(x - 3))(x + 1) \\[0.5em] = (x + 2)(x - 3)(x + 1)$

Hence, the value of a = -7 and b = -6;
x³ - 7x - 6 = (x + 2)(x - 3)(x + 1).

(x - 2) is a factor of the expression x³ + ax² + bx + 6. When this expression is divided by (x - 3), it leaves the remainder 3. Find the values of a and b.

Answer

Let f(x) = x³ + ax² + bx + 6

Given, (x - 2) is factor of f(x), hence, f(2) = 0 by factor's theorem

$\therefore 2^3 + a(2)^2 + b(2) + 6 = 0 \\[0.5em] \Rightarrow 8 + 4a + 2b + 6 = 0 \\[0.5em] \Rightarrow 4a + 2b + 14 = 0 \\[0.5em] \Rightarrow 2a + b + 7 = 0 \\[0.5em] b = -7 - 2a \text{ (Equation 1) }$

Given, on dividing f(x) by (x - 3) remainder left is 3

By remainder theorem, remainder = f(3)

$\therefore 3^3 + a(3)^2 + b(3) + 6 = 3 \\[0.5em] \Rightarrow 27 + 9a + 3b + 6 = 3 \\[0.5em] \Rightarrow 33 + 9a + 3b = 3$

On dividing equation by 3,

$\Rightarrow 11 + 3a + b = 1$

Putting value of b from equation 1,

$\Rightarrow 11 + 3a - 7 - 2a = 1 \\[0.5em] \Rightarrow a + 4 = 1 \\[0.5em] \Rightarrow a = -3 \\[0.5em] \text{ and } b = -7 - 2a = -7 - 2(-3) = -7 + 6 = -1$

Hence, the value of a = -3 and b = -1.

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.

Answer

Let f(x) = 2x³ + ax² + bx - 14

Given, (x - 2) is factor of f(x), hence, f(2) = 0 by factor's theorem

$\therefore 2(2)^3 + a(2)^2 + b(2) - 14 = 0 \\[0.5em] \Rightarrow 16 + 4a + 2b - 14 = 0 \\[0.5em] \Rightarrow 2 + 4a + 2b = 0$

On dividing equation by 2,

$\Rightarrow 2a + b + 1 = 0 \\[0.5em] \Rightarrow b = -1 - 2a \text{ (Equation 1) }$

Given, on dividing f(x) by (x - 3) remainder left is 52

By remainder theorem, remainder = f(3)

$\therefore 2(3)^3 + a(3)^2 + b(3) - 14 = 52 \\[0.5em] \Rightarrow 54 + 9a + 3b - 14 = 52 \\[0.5em] \Rightarrow 9a + 3b + 40 = 52 \\[0.5em] \Rightarrow 9a + 3b = 12 \\[0.5em] \Rightarrow 3a + b = 4$

Putting value of b from equation 1,

$\Rightarrow 3a - 1 - 2a = 4 \\[0.5em] \Rightarrow a = 5 \\[0.5em] \text{ and } b = -1 - 2a = -1 - 10 = -11$

Hence, the value of a = 5 and b = -11.

If ax³ + 3x² + bx - 3 has a factor (2x + 3) and leaves remainder -3 when divided by (x + 2), find the values of a and b. With these values of a and b, factorise the given expression.

Answer

Let f(x) = ax³ + 3x² + bx - 3

Given, (2x + 3) or $2(x - \big(-\dfrac{3}{2}\big))$ is factor of f(x), hence, f$\big(-\dfrac{3}{2}\big)$ = 0 by factor's theorem

$\therefore a\big(-\dfrac{3}{2}\big)^3 + 3\big(-\dfrac{3}{2}\big)^2 + b\big(-\dfrac{3}{2}\big) - 3 = 0 \\[1em] \Rightarrow -\dfrac{27a}{8} + \dfrac{27}{4} - \dfrac{3b}{2} - 3 = 0 \\[1em] \Rightarrow \big(\dfrac{-27a + 54 - 12b - 24}{8}\big) = 0$

On cross multiplication,

$\Rightarrow -27a - 12b + 30 = 0$

On dividing the equation by 3,

$\Rightarrow -9a - 4b + 10 = 0 \\[1em] \Rightarrow 4b + 9a = 10 \text{( Equation 1)}$

Given, on dividing f(x) by (x + 2) remainder left is -3

By remainder theorem, remainder = f(-2)

$\therefore a(-2)^3 + 3(-2)^2 + b(-2) - 3 = -3 \\[0.5em] \Rightarrow -8a + 12 - 2b - 3 = -3 \\[0.5em] \Rightarrow -8a - 2b + 9 = -3 \\[0.5em] \Rightarrow -8a - 2b = -12$

On dividing equation by -2,

$\Rightarrow -4a - b = -6 \\[0.5em] \Rightarrow 4a + b = 6$

Multiplying equation by 4,

$\Rightarrow 16a + 4b = 24$

Subtracting above equation from equation 1,

$\Rightarrow 4b + 9a - 16a - 4b = 10 - 24 \\[0.5em] \Rightarrow -7a = -14 \\[0.5em] \Rightarrow a = 2 \\[0.5em] \text{and } b = 6 - 4a = 6 - 8 = -2.$

Putting value of a and b in f(x) we get,

$f(x) = 2x^3 + 3x^2 - 2x - 3 \\[0.5em] = x^2(2x + 3) -1(2x + 3) \\[0.5em] = (x^2 - 1)(2x + 3) \\[0.5em] = ((x)^2 - (1)^2)(2x + 3) \\[0.5em] = (x - 1)(x + 1)(2x + 3)$

Hence, the value of a = 2 and b = -2 ; 2x³ + 3x² - 2x - 3 = (x - 1)(x + 1)(2x + 3).

Given f(x) = ax² + bx + 2 and g(x) = bx² + ax + 1. If (x - 2) is a factor of f(x) but leaves the remainder -15 when it divides g(x), find the values of a and b. With these values of a and b, factorise the expression

f(x) + g(x) + 4x² + 7x

Answer

f(x) = ax² + bx + 2

Given, (x - 2) is a factor of f(x) hence, by factor theorem f(2) = 0

$\therefore a(2)^2 + b(2) + 2 = 0 \\[0.5em] \Rightarrow 4a + 2b + 2 = 0$

On dividing equation by 2,

$\Rightarrow 2a + b + 1 = 0 \\[0.5em] b = -1 - 2a \text{( Equation 1)}$

g(x) = bx² + ax + 1

Given, on dividing g(x) by (x - 2), remainder = -15 and by remainder theorem, remainder = g(2)

$\therefore g(2) = -15 \\[0.5em] \Rightarrow b(2)^2 + a(2) + 1 = -15 \\[0.5em] \Rightarrow 4b + 2a = -15 - 1 \\[0.5em] \Rightarrow 4b + 2a = -16$

On dividing equation by 2,

$\Rightarrow 2b + a = -8 \\[0.5em]$

Putting value of b = -1 - 2a from equation 1,

$\Rightarrow 2(-1 - 2a) + a = -8 \\[0.5em] \Rightarrow -2 - 4a + a = -8 \\[0.5em] \Rightarrow -3a = -8 + 2 \\[0.5em] \Rightarrow a = \dfrac{-6}{-3} \\[0.5em] \Rightarrow a = 2 \\[0.5em] \text{ and } b = -1 - 2a = -1 - 4 = -5.$

Putting value of a = 2 and b = -5 in f(x) + g(x) + 4x² + 7x we get,

$(2x^2 - 5x + 2) + (-5x^2 + 2x + 1) + 4x^2 + 7x \\[0.5em] = 2x^2 - 5x^2 + 4x^2 - 5x + 2x + 7x + 2 + 1 \\[0.5em] = x^2 + 4x + 3 \\[0.5em] = x^2 + 3x + x + 3 \\[0.5em] = x(x + 3) + 1(x + 3) \\[0.5em] = (x + 1)(x + 3)$

Hence, the value of a = 2 and b = -5;
f(x) + g(x) + 4x² + 7x = (x + 1)(x + 3).