Answer

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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.

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 ${x^2} + 3ax - 2a = 0$.

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 ${x^2} + 3ax - 2a = 0$.

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