Question

Find the values of p and q such that $f\left(x\right)=x^4-p{x}^2+q$
has a factor $x^2-3x+2$

Answer 1

Given that,

$f\left(x\right)=x^4-p{x}^2+q$ has a factor $x^2-3x+2$

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

i.e., $(x-1)$ and $(x-2)$ are the factors of $f\left(x\right)$

By factor theorem, $f\left(1\right)=0$ and $f\left(2\right)=0$

$⟹1^4-p(1{)}^2+q=0$ and $2^4-p(2{)}^2+q=0$

$⟹1-p+q=0$ and $16-4p+q=0$

$p-q=1$………….(I)

and $4p-q=16$……………….(ii)

By solving (I) & (ii) we get,

$p=5$ and $q=4$