Question

Find the value of $p$ and $q$ such that $(x+1)$ and $(x+2)$ are the factors of $x^4+p{x}^3+2x^2-3x+q$.

Answer 1

$P\left(x\right)=x+1=0$

$x=-1$

Substituting,

$P(-1)=(-1{)}^4+p(-1{)}^3+2(-1{)}^2-3(-1)+q=0$

$1-p+2+3+q=0$

$-p+q+6=0$ ------ (i)

$P\left(x\right)=x+2=0$

$x=-2$

$P(-2)=(-2{)}^4+p(-2{)}^3+2(-2{)}^2-3(-2)+q=0$

$16-8p+8+6+q=0$

$-8p+q+30=0$ ------ (ii)

(ii) - (i) gives,

$(-8p+q+30)-(-p+q+6)=0$

$-8p+p+q-q+30-6=0$

$-7p+24=0$

$-7p=-24$

$p=\frac{24}{7}$

Substitute for $p$ in (i)

$-p+q+6=0$

$-\frac{24}{7}+q+6=0$

$q=\frac{24}{7}-6=\frac{24-42}{7}$

$q=\frac{-18}{7}$