Question

Help with one Algebra problem:

Given the polinomial p(x) = (ab+ac-3)x^2 + (ac+bc-6)x+ (ab+bc-9)

And knowing that the ecuation p(x) = 0 has more than two real solutions, find:
abc(a+b)(a+c)(b+c)

I´ve tried to solve p(x)=0 , but looks useless to me.
Any idea ? Thanks.

Answer 1

First note that any quadratic equation can have at most 2 roots (by either the Fundamental Theorem of Algebra, or by using the quadratic formula). Since p(x)=0 has more than 2 roots, it follows that all the coefficients must be zero. This yields us the equations:\[ab+ac-3=0\Longrightarrow ab+ac=3\]\[ac+bc-6=0\Longrightarrow ac+bc=6\]\[ab+bc-9=0\Longrightarrow ab+bc=9\]Try to take it from here, its just formal manipulations from here on out.

Answer 2

Awesome, how i didn´t get that ?.
Now solve that sistem for a, b and c is almost impossible, but, the question is asking us for abc(a+b)(a+c)(b+c) so

\[a(b+c)=3\]
\[c(a+b)=6\]
\[b(a+c)=9\]

Then
\[abc= \frac{ 3}{ b+c }*\frac{ 6}{ a+b }*\frac{ 9 }{ a+c }\]

And
abc(a+b)(a+c)(b+c)= 3*6*9 = 162. !!!

Thank you.