Factorisation

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).

When 2x³ - x² - 3x + 5 is divided by 2x + 1, then the remainder is

1. 6
2. -6
3. -3
4. 0

Answer

By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = 2x³ - x² - 3x + 5

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

$f\big(-\dfrac{1}{2}\big) = 2\big(-\dfrac{1}{2}\big)^3 - \big(-\dfrac{1}{2}\big)^2 - 3\big(-\dfrac{1}{2}\big) + 5 \\[1em] = 2\big(-\dfrac{1}{8}\big) - \dfrac{1}{4} + \dfrac{3}{2} + 5 \\[1em] = -\dfrac{1}{4} - \dfrac{1}{4} + \dfrac{3}{2} + 5 \\[1em] = \dfrac{-1 -1 + 6 + 20}{4} \\[1em] = \dfrac{24}{4} \\[1em] = 6$

∴ Option 1, is the correct option.

If on dividing 4x² - 3kx + 5 by x + 2, the remainder is -3 then the value of k is

1. 4
2. -4
3. 3
4. -3

Answer

By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = 4x² - 3kx + 5

∴ On dividing f(x) by x + 2 or (x - (-2)), Remainder = f(-2)

Given, remainder = -3

∴ f(-2) = -3

$\Rightarrow 4(-2)^2 - 3k(-2) + 5 = -3 \\[0.5em] \Rightarrow 16 + 6k + 5 = -3 \\[0.5em] \Rightarrow 6k + 21 = -3 \\[0.5em] \Rightarrow 6k = -24 \\[0.5em] k = -4$

∴ Option 2, is the correct option.

If on dividing 2x³ + 6x² - (2k - 7)x + 5 by (x + 3), the remainder is k - 1 then the value of k is

1. 2
2. -2
3. -3
4. 3

Answer

By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = 2x³ + 6x² - (2k - 7)x + 5

∴ On dividing f(x) by x + 3 or (x - (-3)), Remainder = f(-3)

Given, remainder = k - 1

∴ f(-3) = k - 1

$\Rightarrow 2(-3)^3 + 6(-3)^2 - (2k - 7)(-3) + 5 = k - 1 \\[0.5em] \Rightarrow -54 + 54 + 6k - 21 + 5 = k - 1 \\[0.5em] \Rightarrow 6k - 16 = k - 1 \\[0.5em] \Rightarrow 6k - k = 16 - 1 \\[0.5em] \Rightarrow 5k = 15 \\[0.5em] \Rightarrow k = 3.$

∴ Option 4, is the correct option.

If x + 1 is a factor of 3x³ + kx² + 7x + 4, then the value of k is

1. -1
2. 0
3. 6
4. 10

Answer

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

f(x) = 3x³ + kx² + 7x + 4

Since, (x + 1) or (x - (-1)) is a factor of f(x),

∴ f(-1) = 0

$\Rightarrow 3(-1)^3 + k(-1)^2 + 7(-1) + 4 = 0 \\[0.5em] \Rightarrow -3 + k - 7 + 4 = 0 \\[0.5em] \Rightarrow k - 6 = 0 \\[0.5em] k = 6$

∴ Option 3, is the correct option.

What must be subtracted from the polynomial x³ + x² - 2x + 1, so that the result is exactly divisible by (x - 3)?

1. –31

2. –30

3. 30

4. 31

Answer

Polynomial : x³ + x² - 2x + 1

Division by x - 3

⇒ x - 3 = 0

⇒ x = 3.

Let k be subtracted from the polynomial, so resulting polynomial is x³ + x² - 2x + 1 - k.

Resulting polynomial should be exactly divisible by x - 3,

Thus, substituting x = 3, in polynomial x³ + x² - 2x + 1 - k, remainder = 0.

⇒ 3³ + 3² - 2(3) + 1 - k = 0

⇒ 27 + 9 - 6 + 1 - k = 0

⇒ 31 - k = 0

⇒ k = 31.

Hence, Option 4 is the correct option.

A polynomial in 'x' is divided by (x - a) and for (x - a) to be a factor of this polynomial, the remainder should be :

1. -a

2. 0

3. a

4. 2a

Answer

For (x - a) to be a factor of a polynomial, the remainder should be equal to zero.

Hence, Option 2 is the correct option.

Given f(x) = ax² + bx + c, a ≠ 0, b, c ∈ R.

Assertion (A): The factorisation of ax² + bx + c is possible only if its discriminant = b² - 4ac < 0.

Reason (R): To factorise ax² + bx + c, split the coefficient of x into two real numbers such that their algebraic sum is b and their product is ac.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given f(x) = ax² + bx + c, a ≠ 0, b, c ∈ R.

Discriminant (D) = b² - 4ac

We know that,

For D ≥ 0, the equation has real roots so factorisation is possible.

So, assertion (A) is false.

To factorise ax² + bx + c, split the coefficient of x into two real numbers such that their algebric sum is b and their product is ac.

This describes the method of splitting the middle term, which is a valid technique used to factor quadratic expressions over the real numbers — provided real roots exist.

So, reason (R) is true.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Given a polynomial f(x) = 2x³ - 7x² - 5x + 4

Assertion (A): (x - 1) is not factor of f(x).

Reason (R): f(1) = -6.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given a polynomial f(x) = 2x³ - 7x² - 5x + 4

By the Factor Theorem, (x - r) is a factor of f(x) if and only if f(r) = 0.

(x - 1) is a factor of f(x) if and only if f(1) = 0.

f(1) = 2 . (1)³ - 7 . (1)² - 5 . (1) + 4

= 2 - 7 - 5 + 4 = -6.

Thus, (x - 1) is not a factor of f(x).

So, Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given f(x) = 16x³ - 8x² + 4x + 7

Assertion (A): When we subtract 1 from f(x), the resulting polynomial is divisible by (2x + 1).

Reason (R): $f\Big(-\dfrac{1}{2}\Big)$ = 1.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given f(x) = 16x³ - 8x² + 4x + 7

Subtract 1 from f(x), we get :

g(x) = 16x³ - 8x² + 4x + 6

By Factor Theorem,

(x - r) is a factor of f(x) if and only if f(r) = 0.

(2x + 1) is a factor of g(x) if and only if $g\Big(-\dfrac{1}{2}\Big)$ = 0.

$g\Big(-\dfrac{1}{2}\Big) = 16 \times \Big(-\dfrac{1}{2}\Big)^3 - 8 \times \Big(-\dfrac{1}{2}\Big)^2 + 4 \times \Big(-\dfrac{1}{2}\Big) + 6\\[1em] = 16 \times \Big(-\dfrac{1}{8}\Big) - 8 \times \Big(\dfrac{1}{4}\Big) + 4 \times \Big(-\dfrac{1}{2}\Big) + 6\\[1em] = -2 - 2 - 2 + 6\\[1em] = 0.$