## RD Sharma class 9 maths Solutions Factorization of Polynomials Exercise-6.4

- Factorization of Polynomials Chapter 6 Ex 6.1
- Factorization of Polynomials Chapter 6 Ex 6.2
- Factorization of Polynomials Chapter 6 Ex 6.3
- Factorization of Polynomials Chapter 6 Ex 6.4
- Factorization of Polynomials Chapter 6 Ex 6.5

### RD Sharma Class 9 Solution Chapter 6 Factorisation of Polynomials Ex 6.4 Latest Edition

In each of the following, use factor Theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1-7)

Question 1.

f(x) = x^{3} – 6x^{2} + 11x – 6; g(x) = x – 3

Solution:

We know that if g(x) is a factor of p(x),

then the remainder will be zero. Now,

f(x) = x^{3} – 6x^{2} + 11x – 6; g(x) = x -3

Let x – 3 = 0, then x = 3

∴ Remainder = f(3)

= (3)^{3} – 6(3)^{2} +11 x 3 – 6

= 27-54 + 33 -6

= 60 – 60 – 0

∵ Remainder is zero,

∴ x – 3 is a factor of f(x)

Question 2.

f(x) = 3X^{4} + 17x^{3} + 9x^{2} – 7x – 10; g(x) = x + 5

Solution:

f(x) = 3x^{4} + 17X^{3} + 9x^{2} – 7x – 10; g(x) = x + 5

Let x + 5 = 0, then x = -5

∴ Remainder = f(-5) = 3(-5)^{4} + 17(-5)^{3} + 9(-5)^{2} – 7(-5) – 10

= 3 x 625 + 17 x (-125) + 9 x (25) – 7 x (-5) – 10

= 1875 -2125 + 225 + 35 – 10

= 2135 – 2135 = 0

∵ Remainder = 0

∴ (x + 5) is a factor of f(x)

Question 3.

f(x) = x^{5} + 3x^{4} – x^{3} – 3x^{2} + 5x + 15, g(x) = x + 3

Solution:

f(x) = x^{5} + 3X^{4} – x^{3} – 3x^{2} + 5x + 15, g(x) = x + 3

Let x + 3 = 0, then x = -3

∴ Remainder = f(-3)

= (-3)^{5} + 3(-3)^{4} – (-3)^{3} – 3(-3)^{2} + 5(-3) + 15

= -243 + 3 x 81 -(-27)-3 x 9 + 5(-3) + 15

= -243 +243 + 27-27- 15 + 15

= 285 – 285 = 0

∵ Remainder = 0

∴ (x + 3) is a factor of f(x)

Question 4.

f(x) = x^{3} – 6x^{2} – 19x + 84, g(x) = x – 7

Solution:

f(x) = x^{3} – 6x^{2} – 19x + 84, g(x) = x – 7

Let x – 7 = 0, then x = 7

∴ Remainder = f(7)

= (7)^{3} – 6(7)^{2} – 19 x 7 + 84

= 343 – 294 – 133 + 84

= 343 + 84 – 294 – 133

= 427 – 427 = 0

∴ Remainder = 0

∴ (x – 7) is a factor of f(x)

Question 5.

f(x) = 3x^{3} + x^{2} – 20x + 12, g(x) = 3x – 2

Solution:

Question 6.

f(x) = 2x^{3} – 9x^{2} + x + 12, g(x) = 3 – 2x

Solution:

f(x) = 2x^{3} – 9x^{2} + x + 12, g(x) = 3 – 2x

Question 7.

f(x) = x^{3} – 6x^{2} + 11x – 6, g(x) = x^{2} – 3x + 2

Solution:

g(x) = x^{2} – 3x + 2

= x^{2} – x – 2x + 2

= x(x – 1) – 2(x – 1)

= (x – 1) (x – 2)

If x – 1 = 0, then x = 1

∴ f(1) = (1)^{3} – 6(1)^{2} + 11(1) – 6

= 1-6+11-6= 12- 12 = 0

∴ Remainder is zero

∴ x – 1 is a factor of f(x)

and if x – 2 = 0, then x = 2

∴ f(2) = (2)^{3} – 6(2)^{2} + 11(2)-6

= 8 – 24 + 22 – 6 = 30 – 30 = 0

∴ Remainder = 0

∴ x – 2 is also a factor of f(x)

Question 8.

Show that (x – 2), (x + 3) and (x – 4) are factors of x^{3} – 3x^{2} – 10x + 24.

Solution:

f(x) = x^{3} – 3x^{2} – 10x + 24

Let x – 2 = 0, then x = 2

Now f(2) = (2)^{3} – 3(2)^{2} – 10 x 2 + 24

= 8 – 12 – 20 + 24 = 32 – 32 = 0

∴ Remainder = 0

∴ (x – 2) is the factor of f(x)

If x + 3 = 0, then x = -3

Now, f(-3) = (-3)^{3} – 3(-3)^{2} – 10 (-3) + 24

= -27 -27 + 30 + 24

= -54 + 54 = 0

∴ Remainder = 0

∴ (x + 3) is a factor of f(x)

If x – 4 = 0, then x = 4

Now f(4) = (4)^{3} – 3(4)^{2} – 10 x 4 + 24 = 64-48 -40 + 24

= 88 – 88 = 0

∴ Remainder = 0

∴ (x – 4) is a factor of (x)

Hence (x – 2), (x + 3) and (x – 4) are the factors of f(x)

Question 9.

Show that (x + 4), (x – 3) and (x – 7) are factors of x^{3} – 6x^{2} – 19x + 84.

Solution:

Let f(x) = x^{3} – 6x^{2} – 19x + 84

If x + 4 = 0, then x = -4

Now, f(-4) = (-4)^{3} – 6(-4)^{2} – 19(-4) + 84

= -64 – 96 + 76 + 84

= 160 – 160 = 0

∴ Remainder = 0

∴ (x + 4) is a factor of f(x)

If x – 3 = 0, then x = 3

Now, f(3) = (3)^{3} – 6(3)^{2} – 19 x 3 + 84

= 27 – 54 – 57 + 84

= 111 -111=0

∴ Remainder = 0

∴ (x – 3) is a factor of f(x)

and if x – 7 = 0, then x = 7

Now, f(7) = (7)^{3} – 6(7)^{2} – 19 x 7 + 84

= 343 – 294 – 133 + 84

= 427 – 427 = 0

∴ Remainder = 0

∴ (x – 7) is also a factor of f(x)

Hence (x + 4), (x – 3), (x – 7) are the factors of f(x)

Question 10.

For what value of a (x – 5) is a factor of x^{3} – 3x^{2} + ax – 10?

Solution:

f(x) = x^{3} – 3x^{2} + ax – 10

Let x – 5 = 0, then x = 5

Now, f(5) = (5)^{3} – 3(5)^{2} + a x 5 – 10

= 125 – 75 + 5a – 10

= 125 – 85 + 5a = 40 + 5a

∴ (x – 5) is a factor of fix)

∴ Remainder = 0

⇒ 40 + 5a = 0 ⇒ 5a = -40

⇒ a = \(\frac { -40 }{ 5 }\)= -8

Hence a = -8

Question 11.

Find the value of a such that (x – 4) is a factor of 5x^{3} – 7x^{2} – ax – 28.

Solution:

Let f(x) = 5x^{3} – 7x^{2} – ax – 28

and Let x – 4 = 0, then x = 4

Now, f(4) = 5(4)^{3} – 7(4)^{2} – a x 4 – 28

= 5 x 64 – 7 x 16 – 4a – 28

= 320 – 112 – 4a – 28

= 320 – 140 – 4a

= 180 – 4a

∴ (x – 4) is a factor of f(x)

∴ Remainder = 0

⇒ 180 -4a = 0

⇒ 4a = 180

⇒ a = \(\frac { 180 }{ 4 }\) = 45

∴ a = 45

Question 12.

Find the value of a, if x + 2 is a factor of 4x^{4} + 2x^{3} – 3x^{2} + 8x + 5a.

Solution:

Let f(x) = 4x^{4} + 2x^{3} – 3x^{2} + 8x + 5a

and Let x + 2 = 0, then x = -2

Now, f(-2) = 4(-2)^{4} + 2(-2)^{3} – 3(-2)^{2} + 8 x ( 2) + 5a

= 4 x 16 + 2(-8) – 3(4) + 8 (-2) + 5a

= 64- 16- 12- 16 +5a

= 64 – 44 + 5a

= 20 + 5a

∴ (x + 2) is a factor of fix)

∴ Remainder = 0

⇒ 20 + 5a = 0 ⇒ 5a = -20

⇒ a =\(\frac { -20 }{ 5 }\) = -4

∴ a = -4

Question 13.

Find the value of k if x – 3 is a factor of k^{2}x^{3} – kx^{2} + 3kx – k.

Solution:

Let f(x) = k^{2}x^{3} – kx^{2} + 3kx – k

and Let x – 3 = 0, then x = 3

Now,f(3) = k^{2}(3)^{3} – k(3)^{2} + 3k(3) – k

= 27k^{2} – 9k + 9k-k

= 27k^{2}-k

∴ x – 3 is a factor of fix)

∴ Remainder = 0

∴ 27k^{2} – k = 0

⇒ k(27k – 1) = 0 Either k = 0

or 21k – 1 = 0

⇒ 21k = 1

∴ k= \(\frac { 1 }{ 27 }\)

∴ k = 0,\(\frac { 1 }{ 27 }\)

Question 14.

Find the values of a and b, if x^{2} – 4 is a factor of ax^{4} + 2x^{3} – 3x^{2} + bx – 4.

Solution:

f(x) = ax^{4} + 2x^{3} – 3x^{2} + bx – 4

Factors of x^{2} – 4 = (x)^{2} – (2)^{2
}= (x + 2) (x – 2)

If x + 2 = 0, then x = -2

Now, f(-2) = a(-2)^{4} + 2(-2)^{3} – 3(-2)^{2} + b(-2) – 4

16a- 16 – 12-26-4

= 16a -2b-32

∵ x + 2 is a factor of f(x)

∴ Remainder = 0

⇒ 16a – 2b – 32 = 0

⇒ 8a – b – 16 = 0

⇒ 8a – b = 16 …(i)

Again x – 2 = 0, then x = 2

Now f(2) = a x (2)^{4} + 2(2)^{3} – 3(2)^{2} + b x 2-4

= 16a + 16- 12 + 26-4

= 16a + 2b

∵ x – 2 is a factor of f(x)

∴ Remainder = 0

⇒ 16a + 2b = 0

⇒ 8a + b= 0 …(ii)

Adding (i) and (ii),

⇒ 16a = 16

⇒ a = \(\frac { 16 }{ 16 }\) = 1

From (ii) 8 x 1 + b = 0

⇒ 8 + b = 0

⇒ b = – 8

∴ a = 1, b = -8

Question 15.

Find α and β, if x + 1 and x + 2 are factors of x^{3} + 3x^{2} – 2αx +β.

Solution:

Let f(x) = x^{3} + 3x^{2} – 2αx + β

and Let x + 1 = 0 then x = -1

Now,f(-1) = (1)^{3} + 3(-1)^{2} – 2α (-1) +β

= -1 + 3 + 2α + β

= 2 + 2α + β

∵ x + 1 is a factor of f(x)

∴ Remainder = 0

∴ 2 + 2α + β = 0

⇒ 2α + β = -2 …(i)

Again, let x + 2 = 0, then x = -2

Now, f(-2) = (-2)^{3} + 3(-2)^{2} – 2α(-2) + β

= -8 + 12 + 4α+ β

= 4 + 4α+ β

∵ x + 2 is a factor of(x)

∴ Remainder = 0

∴ 4+ 4α + β = 0

⇒ 4α + β = -4 …(ii)

Subtracting (i) from (ii),

2α = -2

⇒ α = \(\frac { -2 }{ 2 }\) = -1

From (ii), 4(-1) + β = -4

-4 + β= -4

⇒ β =-4+ 4 = 0

∴ α = -1, β = 0

Question 16.

If x – 2 is a factor of each of the following two polynomials, find the values of a in each case:

(i) x^{3} – 2ax^{2} + ax – 1

(ii) x^{5} – 3x^{4} – ax^{3} + 3ax^{2} + 2ax + 4

Solution:

(i) Let f(x) = x^{3} – 2ax^{2} + ax – 1 and g(x) = x – 2

and let x – 2 = 0, then x = 2

∴ x – 2 is its factor

∴ Remainder = 0

f(2) = (2)^{3} – 2a x (2)^{2} + a x 2 – 1

= 8-8a+ 2a-1 = 7-6a

∴ 7 – 6a = 0

⇒ 6a = 7

⇒ a = \(\frac { 7 }{ 6 }\)

∴ a = \(\frac { 7 }{ 6 }\)

(ii) Let f(x) = x^{5} – 3x^{4} – ax^{3} + 3 ax^{2} + 2ax + 4 and g(x) = x – 2

Let x – 2 = 0, then x=2

∴ f(2) = (2)^{5} – 3(2)^{4} – a(2^{3}) + 3a (2)^{2} + 2a x 2 + 4

= 32 – 48 – 8a + 12a + 4a + 4

= -12 + 8a

∴ Remainder = 0

∴ -12 + 8a = 0

⇒ 8a= 12

⇒ a = \(\frac { 12 }{ 8 }\) = \(\frac { 3 }{ 2 }\)

∴ Hence a = \(\frac { 3 }{ 2 }\)

Question 17.

In each of the following two polynomials, find the values of a, if x – a is a factor:

(i) x^{6} – ax^{5} + x^{4}-ax^{3} + 3x-a + 2

(ii) x^{5} – a^{2}x^{3} + 2x + a + 1

Solution:

(i) Let f(x) ^{=} x^{6 }– ax^{5}+x^{4}-ax^{3} + 3x-a + 2 and g(x) = x – a

∴ x – a is a factor

∴ x – a = 0

⇒ x = a

Now f(a) = a^{6}-a x a^{5} + a^{4}-a x a^{3} + 3a – a + 2

= a^{6}-a^{6} + a^{4}-a^{4} + 2a + 2

= 2a + 2

∴ x + a is a factor of p(x)

∴ Remainder = 0

Question 18.

In each of the following, two polynomials, find the value of a, if x + a is a factor.

(i) x^{3} + ax^{2} – 2x + a + 4

(ii) x^{4} – a^{2}r + 3x – a

Solution:

Question 19.

Find the values of p and q so that x^{4} + px^{3 }+ 2x^{2} – 3x + q is divisible by (x^{2} – 1).

Solution:

Question 20.

Find the values of a and b so that (x + 1) and (x – 1) are factors of x^{4} + ax^{3} 3x^{2} + 2x + b.

Solution:

Question 21.

If x^{3} + ax^{2} – bx + 10 is divisible by x^{2} – 3x + 2, find the values of a and b.

Solution:

Question 22.

If both x + 1 and x – 1 are factors of ax^{3} + x^{2} – 2x + b, find the values of a and b.

Solution:

Question 23.

What must be added to x^{3} – 3x^{2} – 12x + 19 so that the result is exactly divisibly by x^{2} + x – 6?

Solution:

Question 24.

What must be subtracted from x^{3} – 6x^{2} – 15x + 80-so that the result is exactly divisible by x^{2} + x – 12?

Solution:

Question 25.

What must be added to 3x^{3} + x^{2} – 22x + 9 so that the result is exactly divisible by 3x^{2 }+ 7x – 6?

Solution:

### RD Sharma class 9 maths Solutions Factorization of Polynomials Exercise-6.4

RD Sharma class 9 maths Solutions Factorization of Polynomials Exercise-6.4 Q 1.

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