Question

If $x^3+3p{x}^2+3qx+r$ and $x^2+2px+q$ have a common factor, show that
$4(p^2-q)(q^2-pr)-(pq-r{)}^2=0$.
If they have two common factors, show that
$p^2-q=0,q^2-pr=0$.

Answer 1

The required condition may be obtain by eliminating 'x' between the two equations

We have $p{x}^2+2qx+r=0$ and $x^2+2px+1=0$

By cross multiplication;-

$\frac{x^2}{2(q^2-pr)}=\frac{x}{r-pq}=\frac{1}{2(p^2-q)}$

$\therefore 4(q^2-pr)(p^2-q)={(pq-r)}^2$

According to second supposition, the first expression is divisible by the second without remainder

Now,

$x^3+3p{x}^2+3qx+r=(x+p)(x^2+2px+q)+(2q-2p^2)q+r-pq$

Hence,

$2(q-p^2)x+r-pq=0$ for all values of r

$\therefore p^2-q=0$ & $r-pq=0$

$q^2-pr=0(\because pr-p^{2}q=0)$

$\therefore pr=p^{2}q=q^2$