Solve x^3y-8x^2y+12xy

$$x ^ { 3 } y - 8 x ^ { 2 } y + 12 x y$$

Factor

$$xy\left(x-6\right)\left(x-2\right)$$

Solution Steps

Factor out xy.

$$xy\left(x^{2}-8x+12\right)$$

Consider x^{2}-8x+12. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite x^{2}-8x+12 as \left(x^{2}-6x\right)+\left(-2x+12\right).

$$\left(x^{2}-6x\right)+\left(-2x+12\right)$$

Factor out x in the first and -2 in the second group.

$$x\left(x-6\right)-2\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x-2\right)$$

Rewrite the complete factored expression.

$$xy\left(x-6\right)\left(x-2\right)$$

Evaluate

$$xy\left(x-6\right)\left(x-2\right)$$