CEMC Fermat, 2012

Solution 1

Since the two terms have a common factor, then we factor and obtain $(x-2)((x-4)+(x-6))=0$.

This gives $(x-2)(2x-10)=0$.

Therefore, $x-2=0$ (which gives $x=2$) or $2x-10=0$ (which gives $x=5$).

Therefore, the two roots of the equation are $x=2$ and $x=5$. Their product is 10.

Solution 2

We expand and then simplify the left side: $$\begin{aligned} (x-4)(x-2) + (x-2)(x-6) & = & 0 \\ (x^2-6x+8) + (x^2 - 8x + 12) & = & 0 \\ 2x^2 - 14x + 20 & = & 0 \end{aligned}$$ Since the product of the roots of a quadratic equation of the form $ax^2+bx+c=0$ with $a \neq 0$ is $\dfrac{c}{a}$, then the product of the roots of the equation $2x^2-14x+20 =0$ is $\dfrac{20}{2}=10$.