Question

If $x^2-3x+2$ is a factor of $f\left(x\right)=x^4-p{x}^2+q$ ,then $(p,q)=$

• A. $(-4,-5)$
• B. $(4,-5)$
• C. $(-5,-4)$
• D. $(5,4)$

Answer 1

The correct option is D $(5,4)$

Let us first factorize $x^2-3x+2$ as follows:

$x^2-3x+2=x^2-x-2x+2=x(x-1)-2(x-1)=(x-1)(x-2)$

It is given that $x^2-3x+2$ is a factor of the polynomial $f\left(x\right)=x^4-p{x}^2+q$ that is $(x-1)(x-2)$ are the factors of $f\left(x\right)=x^4-p{x}^2+q$ and therefore, $x=1$ and $x=2$ are the zeroes of $f\left(x\right)$.

Now, we substitute $x=1$ and $x=2$ in $f\left(x\right)=x^4-p{x}^2+q$ as shown below:

$f\left(1\right)=1^4-p(1{)}^2+q\Rightarrow 0=1-p+q\Rightarrow p-q=1....(1)$

$f\left(2\right)=2^4-p(2{)}^2+q\Rightarrow 0=16-4p+q\Rightarrow 4p-q=16....(2)$

Subtract equation 1 from equation 2:

$(4p-p)+(-q+q)=16-1\Rightarrow 3p=15\Rightarrow p=\frac{15}{3}\Rightarrow p=5$

Substituting the value of $p$ in equation 1, we get:

$5-q=1\Rightarrow q=5-1\Rightarrow q=4$

Hence, $(p,q)=(5,4)$.