Question

If the polynomial f(x)=ax³+bx-cis divisible by g(x)=x²+bx+c find ab=?

Answer 1

Answer : 1

$f_{(x)}=ax^3+bx-c$

$g_{(x)}=x^2+bx+c$

Since the division gives no remainder

$f_{(x)}\div g_{(x)}=Q_{(x)}...0$

If we do the division

Quotient, here $Q_{(x)}$ : $ax-b$

Remainder, here 0 : $(ab^2-ac+b)x+abc-c$

Since the remainder is 0

It is an identity

$\left \{ {{ab^2+ac+b=0} \atop {abc-c=0}} \right.$

Solve the equation

$c=abc$

If we let c≠0

∴$ab=1$