## RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.3

Other Exercises

- RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1
- RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2
- RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.3
- RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.4
- RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5
- RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials VSAQS
- RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS

In each of the following, using the remainder Theorem, find the remainder when f(x) is divided by g(x) and verify the result by actual division (1 – 8) :

Question 1.

f(x) = x^{3} + 4x^{2} – 3x + 10, g(x) = x + 4

Solution:

Question 2.

f(x) – 4x^{4} – 3x^{3} – 2x^{2}* + *x – 7, g(x)* = *x* – **1
*Solution:

Question 3.

f(x) = 2x^{4} – 6X^{3} + 2x^{2} – x + 2, ,g(x) = x + 2

Solution:

Question 4.

f(x) = 4x^{3} – 12x^{2} + 14x – 3, g(x) = 2x – 1

Solution:

Question 5.

f(x) = x^{3} – 6x^{2} + 2x – 4, g(x) = 1 – 2x

Solution:

Question 6.

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

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

If the polynomials 2x^{3} + ax^{2} + 3x – 5 and x^{3 }+ x^{2} – 4x + a leave the same remainder when divided by x – 2, find the value of a.

Solution:

Let f(x) = 2x^{3} + ax^{2} + 3x – 5

g(x) = x^{3}+x^{2}-4x + a

q(x) = x – 2 ⇒ x-2 = 0 ⇒x = 2

∴ Remainder =f(2) = 2(2)^{3} + a(2)^{2} + 3 x 2-5

= 2 x 8 4-a x 4 + 3 x 2-5

= 16 + 4a + 6 – 5

= 4a +17

and g(2) = (2)^{3} + (2)^{2} -4×2 + a

= 8 + 4 – 8 + a = a + 4

∵ In both cases, remainder are same

∴ 4a + 17 = a + 4

⇒ 4a – a = 4 – 17 ⇒ 3a = -13

⇒ a = \(\frac { -13 }{ 3 }\)

Hence a = \(\frac { -13 }{ 3 }\)

Question 10.

If the polynomials ax^{3} + 3x^{2} – 13 and 2x^{3} – 5x + a, when divided by (x – 2), leave the same remainders, find the value of a.

Solution:

Let p(x) = ax^{3} + 3x^{2} – 13

q(x) = 2x^{3 }–5x + a

and divisor g(x) = x – 2

x-2 = 0

⇒ x = 2

∴ Remainder = p(2) = a(2)^{3} + 3(2)^{2} – 13

= 8a + 12 – 13 = 8a – 1

and q( 2) = 2(2)^{3} – 5×2 + a=16-10 + a

= 6 + a

∵ In each case remainder is same

∴ 8a – 1 = 6 + a

8a – a = 6 + 1

⇒ 7a = 7

⇒ a = \(\frac { 7 }{ 7 }\)= 1

∴ a = 1

Question 11.

Find the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by

Solution:

Question 12.

The polynomials ax^{3} + 3a-^{2} – 3 and 2x^{3} – 5x + a when divided by (x – 4) leave the remainders R_{1} and R_{2}, respectively. Find the values of a in each case of the following cases, if

(i) R_{1} = R_{2}

(ii) R_{1} + R_{2} = 0

(iii) 2R_{1} – R_{2} = 0.

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.3 are helpful to complete your math homework.

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