Question

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

• A. $\frac{2}{3}$
• B. $-\frac{13}{3}$
• C. $\frac{10}{3}$
• D. None of the above

Answer 1

Answer: (B)

$-\frac{13}{3}$

Given polynomial $2x^3+a{x}^2+3x-5$ and $x^3+x^2-4x+a$ divided by x-2 the remainder is same

Then x+2= or x=2 replace x by 2 weget

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

$p\left(2\right)=2(2{)}^3+a(2{)}^2+3\left(2\right)-5$

$\Rightarrow p\left(2\right)=2\times 8+a\times 4+3\times 2-5$

$\Rightarrow P\left(2\right)=16+4a+6-5$

$\Rightarrow p\left(2\right)=4a+17$

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

$\Rightarrow q\left(2\right)=(2{)}^3+(2{)}^2-4\left(2\right)+a$

$\Rightarrow q\left(2\right)=8+4-8+a$

$\Rightarrow A\left(2\right)=4+a$

$\therefore 4a+17=4+a$

Add -a and -17 both side we get

$4a+17-a-17=4+a-a-17$

$\Rightarrow 3a=-13$

Divided both side by 3 we get

$a=-\frac{13}{3}$