RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.3 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 3.3.
| Board | RBSE |
| Class | Class 9 |
| Subject | Maths |
| Chapter | Chapter 3 |
| Chapter Name | Polynomial |
| Exercise | Ex 3.3 |
| Number of Questions Solved | 4 |
| Category | RBSE Solutions |
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.3
Question 1.
Find the remainder when dividing (RBSESolutions.com) the polynomial x4 + x3 – 3x2 + 3x + 1 by each of the following one degree expression.
(i) x – 1
(ii) \(x-\frac { 1 }{ 2 }\)
(iii) x + π
(iv) 3 + 2x
(v) x
Solution.
(i) Let p(x) = x4 + x3 – 3x2 + 3x + 1
zero of x – 1 is 1
∴p(1) = (1)4 + (1)3 – 3(1) + 1
= 1 + 1 – 3 + 3 + 1 = 3
Question 2.
Find the remainder when 2x3 + 20x2 – 5x + a is (RBSESolutions.com) divided by x + a.
Solution.
Let p(x) = 2x3 + 2ax2 – 5x + a
Zero of x + a is x = -a
∴ p(- a) = 2(- a)3 + 2a(- a)2 – 5(- a) + a
= – 2a3 + 2a3 + 5a + a = 6a
Question 3.
Verify whether x + 1 is (RBSESolutions.com) a factor of x3 + 3x2 + 3x + 1 or not.
Solution.
Let p(x) = x3 + 3x2 + 3x + 1
If x + 1 is a factor of p(x), then p(- 1) = 0
∴ p(- 1) = (- 1)3 + 3(- 1)2 + 3(- 1) + 1
= – 1 + 3 – 3 + 1 = 0
∵ p(- 1) = 0
∴ x + 1 is a factor of p(x).
Question 4.
Polynomial x3 + x2 – 4x + a and 2x3 + ax2 + 3x – 3 when divided by x – 2 gives (RBSESolutions.com) same remainder. Find the value of a.
Solution.
Let p(x) = x3 + x2 – 4x + a
Here, p(x) is divided by x – 2, so remainder is p(2).
i. e. p(2) = (2)3 + (2)2 – 4(2) + a
⇒ p( 2) = 8 + 4 – 8 + a = a + 4
Again let f(x) = 2x3 + ax2 + 3x – 3 is (RBSESolutions.com) divided by x – 2.
So remainder = f(2)
∴ f(2) = 2(2)3 + a(2)2 + 3(2) – 3
= 16 + 4a + 6 – 3 = 4a + 19
∵ p( 2) = f(2)
⇒ a + 4 = 4a + 19
⇒ 3a = – 15
⇒ a = – 5
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