The product of the roots of the equation $(x-4)(x-2)+(x-2)(x-6)=0$ is
(A) 20
(B) 48
(C) 10
(D) 96
(E) 2
Solution 1

Since the two terms have a common factor, then we factor and obtain $(x-2)((x-4)+(x-6))=0$. This gives $(x-2)(2 x-10)=0$.

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

$$
\begin{array}{r}
(x-4)(x-2)+(x-2)(x-6)=0 \\
\left(x^{2}-6 x+8\right)+\left(x^{2}-8 x+12\right)=0 \\
2 x^{2}-14 x+20=0
\end{array}
$$

Since the product of the roots of a quadratic equation of the form $a x^{2}+b x+c=0$ with $a \neq 0$ is $\frac{c}{a}$, then the product of the roots of the equation $2 x^{2}-14 x+20=0$ is $\frac{20}{2}=10$.

ANSWER: (C)