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# Find the value of $a$ if $x - 2$ is a factor of ${x^2} + 3ax - 2a$.(A) 2 (B) -2 (C) 1 (D) -1 Verified
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Hint: If $x - a$ is a factor of any polynomial $f\left( x \right)$, then $f\left( a \right) = 0$ i.e. while putting $x = a$ in the polynomial we will get zero as a result. Use this theorem for the given polynomial and its factor.

According to the question, the polynomial given is ${x^2} + 3ax - 2a$. Let it be $f\left( x \right)$. So we have:
$\Rightarrow f\left( x \right) = {x^2} + 3ax - 2a$
And it is mentioned that $x - 2$ is a factor of the polynomial.
According to the factor theorem, we know that if $x - a$ is a factor of any polynomial $f\left( x \right)$, then $f\left( a \right) = 0$.
So, if we put $x = a$ in the polynomial, we will get zero as a final result. We will use this theorem for the polynomial given in the question.
Thus, if $x - 2$ is a factor of $f\left( x \right)$, then $f\left( 2 \right) = 0$. So we have:
$\Rightarrow f\left( x \right) = {x^2} + 3ax - 2a \\ \Rightarrow f\left( 2 \right) = {2^2} + 3a \times 2 - 2a \\ \Rightarrow 0 = 4 + 6a - 2a \\ \Rightarrow 4a + 4 = 0 \\ \Rightarrow 4a = - 4 \\ \Rightarrow a = - 1 \\$
Therefore, the required value of a is -1. (D) is the correct option.

Note: If $x - a$ is a factor of $f\left( x \right)$, then $x = a$ is also a root of $f\left( x \right) = 0$.
In the above question, $x - 2$ is a factor of ${x^2} + 3ax - 2a$. So, $x = 2$ is also a root of ${x^2} + 3ax - 2a = 0$.
Last updated date: 27th Sep 2023
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