# Difference between revisions of "2007 iTest Problems/Problem 23"

## Problem

Find the product of the non-real roots of the equation $$(x^2-3x)^2+5(x^2-3x)+6=0$$

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } -1\quad \text{(D) } 2\quad \text{(E) } -2\quad \text{(F) } 3\quad \text{(G) } -3\quad \text{(H) } 4\quad \text{(I) } -4\quad$

$\text{(J) } 5\quad \text{(K) } -5\quad \text{(L) } 6\quad \text{(M) } -6\quad \text{(N) } 3+2i\quad \text{(O) } 3-2i\quad$

$\text{(P) } \frac{-3+i\sqrt{3}}{2}\quad \text{(Q) } 8\quad \text{(R) } -8\qquad \text{(S) } 12\quad \text{(T) } -12\quad \text{(U) } 42\quad$

$\text{(V) Ying} \quad \text{(W) } 207$

## Solution

Let $a = x^2 - 3x$. Substituting that in results in $$a^2 + 5a + 6$$ Factoring that (and substituting back) leads to $$(a+2)(a+3)$$ $$(x^2-3x+2)(x^2-3x+3)$$ The first part is factorable into $(x-2)(x-1)$. The second part isn’t — and the discriminant is $3^2 - 4 \cdot 3 = -3$. With the imaginary factors found, by Vieta's Formulas, the product of the non-real roots is $\boxed{\textbf{(F) }3}$.