In this chapter, we provide RD Sharma Solutions for Chapter 6 Factorisation of Polynomials Ex 6.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 6 Factorisation of Polynomials Ex 6.3 Maths pdf, free RD Sharma Solutions for Chapter 6 Factorisation of Polynomials Ex 6.3 Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 6 |

Chapter Name | Factorisation of Polynomials |

Exercise | Ex 6.3 |

**RD Sharma Solutions for Class 9 Chapter 6** **Factorisation of Polynomials Ex 6.3 Download PDF**

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 = −133

Hence a = −133

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 = 77= 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:

**All Chapter RD Sharma Solutions For Class 9 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class 9**

**All Subject NCERT Solutions For Class 9**

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