Question

In each of the following two polynomials, find the value of $a$, if $x-a$ is a factor:

$x^6-a{x}^5+x^4-a{x}^3+3x-a+2$

Answer 1

Given: $(x-a)$ is a factor of $x^6-a{x}^5+x^4-a{x}^3+3x-a+2$.

Let, $f\left(x\right)=x^6-a{x}^5+x^4-a{x}^3+3x-a+2$
$\because (x-a)$ is a factor of $x^6-a{x}^5+x^4-a{x}^3+3x-a+2$

$\therefore f\left(a\right)=0$ [ Using Factor theorem]

$\Rightarrow (a{)}^6-a\times (a{)}^5+(a{)}^4-a\times (a{)}^3+3\times a-a+2=0$

$\Rightarrow a^6-a^6+a^4-a^4+3a-a+2=0$

$\Rightarrow 2a+2=0\Rightarrow 2a=-2$

$\therefore a=-1$

Hence, the value of $a=-1$.