Question

By actual division, we get,
[
frac{a x^{3}+b x+c}{x^{2}+p x+1}=a x-a p+frac{R(x)}{x^{2}+p x+1}
]
Remainder polynomial, ( R(x)=left(b-a+a p^{2}right) x+c+a p )
If ( a x^{3}+b x+c ) has a factor of the form ( x^{2}+p x+1, ) then ( R(x) ) must be identicallyzero, if ( b-a+a p^{2}=0 ) and ( c+a p=0 ) Eliminating p from these equations, we get
( b-a+aleft(-frac{c}{a}right)^{2}=0 quad ) or ( quad a^{2}-c^{2}=a b )

# Ifax3 + bx+c has a factor of the form x2 + px + 1, show that a--ca = ab.

Solution