## RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials VSAQS

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

**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

**Question 1.
**

**Define zero or root of a polynomial.**

**Solution:**

A real number a is a zero or root of a polynomial f(x) if f(α) = 0

**Question 2.
**

**If x = \(\frac { 1 }{ 2 }\) is a zero of the polynomial f(x) = 8x**

^{3}+ ax^{2}– 4x + 2, find the value of a.**Solution:**

**Question 3.
**

**Write the remainder when the polynomial f(x) = x**

^{3}+x^{2}-3x + 2is divided by x + 1.**Solution:**

f(x) = x

^{3}+x

^{2}-3x + 2

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

∴ Remainder =(-1)

Now,f(-1) = (-1)

^{3}+ (-1)

^{2}– 3(-1) + 2

= -1 + 1+ 3 + 2 = 5

∴ Remainder = 5

**Question 4.
**

**Find the remainder when x**

^{3}+ 4x^{2}+ 4x – 3 is divided by x.**Solution:**

f(x) = x

^{3}+ 4x

^{2}+ 4x – 3

Dividing f(x) by x, we get

Let x = 0, then

f(x) = 0 + 0 + 0 – 3 = -3

∴ Remainder = -3

**Question 5.
**

**If x + 1 is a factor of x**

^{3}+ a, then write the value of a.**Solution:**

Let f(x) = x

^{3}+ a

∴ x + 1 is a factor of fx)

Let x + 1 = 0

⇒ x = -1

∴ f(-1) = x

^{3}+ a

= (-1)

^{3}+ a = -1 + a

∴ x + 1 is a factor

∴ Remainder = 0

∴ -1 + a = 0 ⇒ a = 1

Hence a = 1

**Question 6.
**

**If f(x) = x**

^{4}-2x^{3}+ 3x^{2}– ax – b when divided by x – 1, the remainder is 6, then find the value of a + b.**Solution:**

f(x) = x

^{4}– 2x

^{3}+ 3x

^{2}– ax – b

Dividing f(x) by x – 1, the remainder = 6

Now let x – 1 = 0, then x = 1

∴ f(1) = (1)

^{4}– 2(1)

^{3}+ 3(1)

^{2}-ax 1-b

= 1 -2 + 3-a-b = 2-a-b

∴ Remainder = 6

∴ 2 – a – b = 6 ⇒ a + b = 2 – 6 = -4

Hence a + b = -4

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

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