Question

If $x-2$ is a factor of each of the following two polynomials, find the value of $a$ in each case

$x^5-3x^4-a{x}^3+3a{x}^2+2ax+4$

Answer 1

Given: $(x-2)$ is a factor of $x^5-3x^4-a{x}^3+3a{x}^2+2ax+4$.

Let, $f\left(x\right)=x^5-3x^4-a{x}^3+3a{x}^2+2ax+4$

$\because (x-2)$ is a factor of
$x^5-3x^4-a{x}^3+3a{x}^2+2ax+4$;

then by factor theorem $f\left(2\right)=0$.

$\because f\left(2\right)=0$

$\Rightarrow (2{)}^5-3\times (2{)}^4-a\times (2{)}^3+3\times a\times (2{)}^2+2a\times \left(2\right)+4=0$

$\Rightarrow 32-48-8a+12a+4a+4=0$

$\Rightarrow 8a-12=0$

$\Rightarrow 8a=12$

$\therefore a=\frac{12}{8}=\frac{3}{2}$

Hence, the value of $a$ is $\frac{3}{2}$.