Question

If the polynomials $2x^2+a{x}^2+3x-5$ and $x^2+x^2+-4x+a$ leave the same remainder when divided by $(x-2)$. Find the value of $a$

• A. -$\frac{13}{3}$
• B. -$\frac{1}{3}$
• C. $\frac{13}{3}$
• D. None of these

Answer 1

The correct option is A -$\frac{13}{3}$

$x-2)2x^3+a{x}^2+3x-5(2x^2+\left(a+4\right)x+2a+1)$

$2x^3-4x^2$

$\underset{–}{+-}$

$\left(a+4\right)x^2+3x$

$\left(a+4\right)x^2-2\left(a+4\right)x$

$\underset{–}{+-}$

$\left(2a+11\right)x-5$

$\left(2a+11\right)x-4a-22$

$\underset{–}{+-}$

$4a+17$

$x-2)x^3+x^2-4x+a(x^2+3x+2$

$x^3-2x^2$

$\underset{–}{+-}$

$3x^2-4x$

$3x^2-6x$

$\underset{–}{+-}$

$2x+a$

$2x-4$

$\underset{–}{+-}$

$4+a$

Now remainder are equal

So $4a+17=4+a$

$\Rightarrow 3a=-13$

$\Rightarrow a=-13/3$