Question

If $x-2$ is the factor of ${{x}^{5}}-3{{x}^{4}}-a{{x}^{3}}+3a{{x}^{2}}+2ax+4$ , find the value of a.

Answer 1

Hint: We need to find the value of the variable a if $x-2$ is the factor of the polynomial${{x}^{5}}-3{{x}^{4}}-a{{x}^{3}}+3a{{x}^{2}}+2ax+4$. We solve the given question by equating the polynomial to zero after substituting x = 2 to get the desired result.

Complete step by step answer:
We are given a polynomial and are asked to find the value of the variable a if $x-2$ is the factor of the polynomial${{x}^{5}}-3{{x}^{4}}-a{{x}^{3}}+3a{{x}^{2}}+2ax+4$.
We will be solving the given question by equating the polynomial to zero after substituting x = 2 to get the value of a as per the concept of factor theorem.
According to our question,
Let the value of $f\left( x \right)={{x}^{5}}-3{{x}^{4}}-a{{x}^{3}}+3a{{x}^{2}}+2ax+4$
The expression $x-2$ is the factor of $f\left( x \right)$
From the concept of factors and multiples,
We know that If p, q are the roots of the polynomial then the factors of the polynomial are given by $\left( x-p \right)\left( x-q \right)$ .
According to the factor theorem,
If the value $f\left( r \right)$ is equal to zero then $x-r$ is considered to be the factor of the polynomial $f\left( x \right)$
From the factor theorem,
We can say that the value of $f\left( 2 \right)$ must be equal to zero as $x-2$ is the factor of $f\left( x \right)$
 $\Rightarrow f\left( x \right)={{x}^{5}}-3{{x}^{4}}-a{{x}^{3}}+3a{{x}^{2}}+2ax+4$
Substituting x = 2 in the above equation, we get,
$\Rightarrow f\left( 2 \right)={{2}^{5}}-3\left( {{2}^{4}} \right)-a\left( {{2}^{3}} \right)+3a\left( {{2}^{2}} \right)+2a\left( 2 \right)+4$
Simplifying the above equation, we get,
$\Rightarrow f\left( 2 \right)=32-3\left( 16 \right)-a\left( 8 \right)+3a\left( 4 \right)+2a\left( 2 \right)+4$
Let us evaluate it further.
$\Rightarrow f\left( 2 \right)=32-48-8a+12a+4a+4$
Performing arithmetic operations on the like terms, we get,
$\Rightarrow f\left( 2 \right)=-12+8a$
From the factor-theorem, we know that the value of $f\left( 2 \right)=0$
Applying the same, we get,
$\Rightarrow 0=-12+8a$
Moving the term -12 to the other side of the equation, we get,
$\Rightarrow 12=8a$
Moving the term 8 to the other side of the equation, we get,
$\Rightarrow \dfrac{12}{8}=a$
Canceling the common factors, we get,
$\Rightarrow \dfrac{3}{2}=a$
Rewriting the above equation, we get,
$\therefore a=\dfrac{3}{2}$
Hence, the value of a is $\dfrac{3}{2}$

Note: In algebra, the like terms are the terms that have the same variable raised to the same power. The coefficients need not match for the terms. We must remember that only the like terms can be added or subtracted in algebra.