Question

If the polynomials ax^3 + 4x^2 + 3x - 5 and x^3 - 3x^2 - 5x + a leave the same remainder when divided by (x-3) and (x+2) respectively, find the value of a.....

Answer 1

Given:

The polynomials $ax^3+4x^2+3x-5$ and $x^3-3x^2-5x+a$ leave the same remainder when divided by (x-3) and (x+2) respectively.

To find:

The value of a.

Solution:

Remainder theorem: If a polynomial p(x) is divided by (x-c), then the remainder is equal to p(c).

The polynomials $f(x)=ax^3+4x^2+3x-5$ is divided by (x-3). So, the remainder is f(3).

$f(3)=a(3)^3+4(3)^2+3(3)-5$

$f(3)=27a+36+9-5$

$f(3)=27a+40$

The polynomial $g(x)=x^3-3x^2-5x+a$ is divided by (x+2). So, the remainder is g(-2).

$g(-2)=(-2)^3-3(-2)^2-5(-2)+a$

$g(-2)=-8-12+10+a$

$g(-2)=-10+a$

It is given that the remainders are same. So,

$f(3)=g(-2)$

$27a+40=-10+a$

$27a-a=-10-40$

$26a=-50$

Divide both sides by 26.

$a=\dfrac{-50}{26}$

$a=\dfrac{-25}{13}$

Therefore, the value of a is $\dfrac{-25}{13}$.