Question

If $x+a$ is a factor of $x^4–a^2{x}^2+3x–6a$, then the value of '$a$' is

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

The correct option is C

$0$

Explanation of the correct option

Substituting the fraction with its factor

Given: $x+a$ is a factor of $x^4–a^2{x}^2+3x–6a$

$\Rightarrow {\left(-a\right)}^4–a^2\times {\left(-a\right)}^2+3\left(-a\right)–6a=0$ $\left[\because x=-a\right]$

$$\begin{gathered} \Rightarrow a^4–a^4-3a-6a=0 \\ \Rightarrow 0-9a=0 \\ \Rightarrow a=0 \end{gathered}$$ [$\because {\left(-a\right)}^x=a^x$$,x$ is even]

Hence, the value of $a$ is equal to $0$

Hence, option C is correct.