Question

Find the value of $a$ if $x - 2$ is a factor of ${x^2} + 3ax - 2a$.
(A) 2
(B) -2
(C) 1
(D) -1

Answer 1

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.

Complete step-by-step answer:
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 ${x^2} + 3ax - 2a = 0$.