If both ( x - 2 ) and x - 1 2 are factors of p x 2 + 5 x + r , prove that p = r .

Consider, f ( x ) = p x 2 + 5 x + r According to the factor theorem x - a is a factor of f x then f a = 0 . So, ( x - 2 ) and x - 1 2 are factors of f ( x ) , if f ( 2 ) and f 1 2 = 0 . Thus, f ( 2 ) = 0 ⇒ p ( 2 ) 2 + 5 ( 2 ) + r = 0 ⇒ 4 p + 10 + r = 0 ⇒ 4 p + r = - 10 _______________ 1 And f 1 2 = 0 ⇒ p 1 2 2 + 5 1 2 + r = 0 ⇒ p 4 + 5 2 + r = 0 ⇒ p + 10 + 4 r 4 = 0 ⇒ p + 10 + 4 r = 0 p + 4 r = - 10 _______________ 2 From equation 1 and 2 , we get, 4 p + r = p + 4 r ⇒ 4 p - p = 4 r - r ⇒ 3 p = 3 r ⇒ p = r Hence, proved.

E M B I B E

Secondary School Mathematics FOR CLASS 9>Algebra>Chapter 2 - Polynomials>EXERCISE 2D

R S Aggarwal and Veena Aggarwal Solutions for Exercise 4: EXERCISE 2D

Author:R S Aggarwal & Veena Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal and Veena Aggarwal Solutions for Exercise 4: EXERCISE 2D

Attempt the free practice questions from Exercise 4: EXERCISE 2D with hints and solutions to strengthen your understanding. Secondary School Mathematics FOR CLASS 9 solutions are prepared by Experienced Embibe Experts.

Questions from R S Aggarwal and Veena Aggarwal Solutions for Exercise 4: EXERCISE 2D with Hints & Solutions

MEDIUM

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 2 - Polynomials>EXERCISE 2D>Q 14

Find the value of $a$ for which $(x + 1)$ is a factor of $(a x^{3} + x^{2} - 2 x + 4 a - 9)$

MEDIUM

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 2 - Polynomials>EXERCISE 2D>Q 15

Find the value of $a$ for which $(x + 2 a)$ is a factor of $(x^{5} - 4 a^{2} x^{3} + 2 x + 2 a + 3)$

EASY

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 2 - Polynomials>EXERCISE 2D>Q 16

Find the value of $m$ for which $(2 x - 1)$ is a factor of $(8 x^{4} + 4 x^{3} - 16 x^{2} + 10 x + m)$ .

EASY

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 2 - Polynomials>EXERCISE 2D>Q 22

If both $(x - 2)$ and $(x - \frac{1}{2})$ are factors of $p x^{2} + 5 x + r$ , prove that $p = r$ .

EASY

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 2 - Polynomials>EXERCISE 2D>Q 23

Without actual division, prove that $2 x^{4} - 5 x^{3} + 2 x^{2} - x + 2$ is divisible by $x^{2} - 3 x + 2$

EASY

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 2 - Polynomials>EXERCISE 2D>Q 24

What must be added to $2 x^{4} - 5 x^{3} + 2 x^{2} - x - 3$ so that the result is exactly divisible by $(x - 2)$ ?

EASY

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 2 - Polynomials>EXERCISE 2D>Q 25

If $(a x + b)$ must be subtracted from $(x^{4} + 2 x^{3} - 2 x^{2} + 4 x + 6)$ so that the result is exactly divisible by $(x^{2} + 2 x - 3)$ , then find the value of $(a + b)$ .