RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.4 is part of RBSE Solutions for Class 9 Maths. Here we have given RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.4.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Polynomial |

Exercise |
Ex 3.4 |

Number of Questions Solved |
7 |

Category |
RBSE Solutions |

## RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.4

Question 1.

Determine which of the following (RBSESolutions.com) polynomials has x – 1 factor.

(i) x^{4} – 2x^{3} – 3x^{2} + 2x + 2

(ii) x^{4} + x^{3} + x^{2} + x + 1

(iii) x^{4} + 3x^{3} – 3x^{2} + x – 2

(iv) x^{3} – x^{2} – (2 + √3)x + √3

Solution.

(i) Let p(x) = x^{4} – 2x^{3} – 3x^{2} + 2x + 2 and

g(x) = x – 1

If x – 1 i.e. g(x) be (RBSESolutions.com) a factor of p(x) then

P(1) = 0

Now p(1) = (1)^{4} – 2(1)^{3} – 3(1)^{2} + 2(1) + 2

= 1 – 2 – 3 + 2 + 2 = 0

∵ p(1) = 0

=> x – 1 is a factor of

p(x) = x^{4} – 2x^{3} – 3x^{2} + 2x + 2.

(ii) Let p(x) = x^{4} + x^{3} + x^{2} + x + 1

If x – 1 is a factor of p(x) then p(1) = 0

p(1) = (1)^{4} + (1)^{3} + (1)^{2} + (1) + 1

= 1 + 1 + 1 + 1 + 1 = 5

p(1) ≠ 0

∴ x – 1 is a factor of x^{4} + x^{3} + x^{2} + x + 1.

(iii) Let p(x) = x^{4} + 3x^{3} – 3x^{2} + x – 2

If x – 1 is (RBSESolutions.com) a factor of p(x) then p(1) = 0

∴ p(1) = (1)^{4} + 3(1)^{3} – 3(1)^{2} + (1) – 2

= 1 + 3 – 3 + 1 – 2 = 0

p(1) = 0

∴ x – 1 is a factor of x^{4} + 3x^{3} – 3x^{2} + x – 2.

(iv) Let p(x) = x^{3} – x^{2} – (2 + √3)x + √3

If x – 1 is a factor of p(x) then p(1) = 0

∴ p(1)= (1)^{3} – (1)^{2} – (2 + √3)(1) + √3

= 1 – 1 – 2 – √3 + √3 = -2

∴ p( 1) ≠ 0

Hence, x – 1 is not a factor of x^{3} – x^{2} – (2 + √3)x + √3

Question 2.

Use the factor theorem (RBSESolutions.com) to determine whether g(x) is a factor of p(x) in each of the following case.

(i) p(x) = 3x^{3} – x^{2} – 3x + 1; g(x) = x + 1

(ii) p(x) = 2x^{4} – 7x^{3} – 13x^{2} + 63x – 45; g(x) = x – 1

(iii) p(x) = 3x^{3} + 3x^{2} + 3x + 1; g(x) = x + 2

(iv) p(x) = 2x^{3} + x^{2} – 2x – 1; g(x) = 2x + 1

Solution.

(i) Here p(x) = 3x^{3} – x^{2} – 3x + 1 and g(x) = x + 1

Now we will try to find p(- 1). If p(- 1) = 0 then we say that g(x) is a factor of p(x) otherwise not.

Now p(- 1) = 3(- 1)^{3} – (- 1)^{2} – 3(- 1) + 1

= 3(- 1) – (1) + 3 + 1

= -3 – 1 + 3 + 1 = 0

p(1) = 0

∴ g(x) is a factor of p(x)

i.e. x + 1 is a factor of 3x^{3} – x^{2} – 3x + 1.

(ii) Here

p(x) = 2x^{4} – 7x^{3} – 13x^{2} + 63x – 45

and g(x) = x – 1

Now

p( 1) = 2(1)^{4} – 7(1)^{3} – 13(1)^{2} + 63(1) – 45

= 2 – 7 – 13 + 63 – 45

= – 65 + 65 = 0

p( 1) = 0

i.e. x – 1 is a factor of 2x^{4} – 7x^{3} – 13x^{2} + 63x – 45

(iii) Here p(x) = 3x^{3} + 3x^{2} + 3x + 1 and g(x) = x + 2

Now

p(- 2) = 3(- 2)^{3} + 3(- 2)^{2} + 3(- 2) + 1 = -24 + 12 – 6 + 1 = – 30 + 13 = -17

P(- 2) ≠ 0

i.e. x + 2 is not a factor of 3x^{3} + 3x^{2} + 3x + 1

(iv) Here p(x) = 2x^{3} + x^{2} – 2x – 1 and

g(x) = x + 1

Now

i.e. 2x + 1 is a factor of 2x^{3} + x^{2} – 2x – 1.

Question 3.

Find the (RBSESolutions.com) value of k, when x – 5 is a factor of x^{3} – 3x^{2} + kx – 10.

Solution.

Let p(x) = x^{3} – 3x^{2} + kx – 10

When x – 5 is a factor of p(x) then P( 5) = 0

⇒ (5)^{3} – 3(5)^{2} + k x 5 – 10 = 0

⇒ 125 – 75 + 5k – 10 = 0

⇒ 5k – 40 = 0

k = \(\frac { 40 }{ 5 }\) = 8

Hence, k = 8

Question 4.

Find the value of k, when x – 1 is a factor of 2x^{2} + kx + √2

Solution.

Let p(x) = 2x^{2} + kx + √2

When x – 1 is a factor of p(x) then p(1) = 0

⇒ 2(1)^{2} + k(1) + √2 = 0

⇒ 2 + k + √2 = 0

⇒ k = -(2 + √2)

Question 5.

If (x + 1) and (x – 1) are the (RBSESolutions.com) factors of x^{4} + ax^{3} – 3x^{2} + 2x + b then find the values of a and b.

Solution.

Let p(x) = x^{4} + ax^{3} – 3x^{2} + 2x + b

∵ (x + 1) is a factor of p(x)

∴ P(- 1) = 0

=> (-1)^{4} + a(-1)^{3} – 3(- 1)^{2} + 2(- 1) + b = 0

=> 1 – a – 3 – 2 + b = 0

=> – a + b = 4 ….(i)

Also (x – 1) is a factor of p(x)

∴ P( 1) = 0

=> (1)^{4} + a(1)^{3} – 3(1)^{2} + 2(1) + b = 0

=> 1 + a – 3 + 2 + 6 = 0

=> a + b = 0 …(ii)

Solving (i) and (ii), we get

2b – 4 => b = 2

Substituting b = 2 in (ii), we get

a + 2 = 0 => a = -2

Hence, a = – 2, b = 2

Question 6.

Factorise by splitting (RBSESolutions.com) middle term

(i) 3x^{2} + 7x + 2

(ii) 4x^{2} – x – 3

(iii)12x^{2} – 7x + 1

(iv) 6x^{2} + 5x – 6

Solution.

(i) Let p(x) = 3x^{2} + 7x + 2

Here we have to split middle term of p(x)

i.e. 7 into two parts such (RBSESolutions.com) that their sum is 7 and product is 2 x 3 = 6.

6 and 1 are such numbers whose sum is 7 and product is 6

∴ 3x^{2} + 7x + 2 = 3x^{2} + 6x + x + 2

= 3x(x + 2) + 1(x + 2)

= (x + 2)(3x + 1)

(ii) Let p(x) = 4x^{2} – x – 3

Here we have to split middle term – 1 into two parts such that their sum is – 1 product is 3 × – 4 = – 12

– 4 and 3 are such numbers whose sum is – 1 and product is – 12

4x^{2} – x – 3 = 4x^{2} – 4x + 3x – 3

= 4x(x – 1) + 3(x – 1)

= (x – 1 )(4x + 3)

(iii) Let p(x) = 12x^{2} – 7x + 1

Here we have to split middle term – 7 into (RBSESolutions.com) two parts such that their sum is – 7 and product is 12 × 1 = 12

– 4 and – 3 are such numbers whose sum is – 7 and product is 12

12x^{2} – 7x + 1 = 12x^{2} – 4x – 3x + 1

= 4x(3x – 1) – 1(3x – 1)

= (3x – 1)(4x – 1)

(iv) Let p(x) = 6x^{2} + 5x – 6

Here we have to split middle term 5 into two parts such that their sum is 5 and product is 6 × – 6 = – 36

9 and – 4 are such numbers whose sum is 5 and product is – 36. .

6x^{2} + 5x – 6 = 6x^{2} + 9x – 5x – 6

= 3x(2x + 3) – 2(2x + 3) = (2x + 3)(3x – 2)

Question 7.

Find the zeroes of (RBSESolutions.com) the polynomials:

(i) x^{3} + 6x^{2} + 11x + 6

(ii) x^{3} + 2x^{2} – x – 2

(iii) x^{4} – 2x^{3} – 7x^{2} + 8x + 12

(iv) x^{3} – 2x^{3} – x + 2

(v) x^{3} – 3x^{2} – 9x – 5

(vi) x^{3} – 23x^{2} + 142x – 120

Solution.

(i) Let the given (RBSESolutions.com) polynomial be

p(x) = x^{3} + 6x^{2} + 11x + 6

The coefficient of the leading term is 1

and the constant term is 6.

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