Solve x^3-13x^2+42x

$$x ^ { 3 } - 13 x ^ { 2 } + 42 x$$

Factor

$$x\left(x-7\right)\left(x-6\right)$$

Solution Steps

Factor out x.

$$x\left(x^{2}-13x+42\right)$$

Consider x^{2}-13x+42. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+42. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 42=42$$

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 42.

$$-1,-42 -2,-21 -3,-14 -6,-7$$

Calculate the sum for each pair.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

The solution is the pair that gives sum -13.

$$a=-7 b=-6$$

Rewrite x^{2}-13x+42 as \left(x^{2}-7x\right)+\left(-6x+42\right).

$$\left(x^{2}-7x\right)+\left(-6x+42\right)$$

Factor out x in the first and -6 in the second group.

$$x\left(x-7\right)-6\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x-6\right)$$

Rewrite the complete factored expression.

$$x\left(x-7\right)\left(x-6\right)$$

Evaluate

$$x\left(x-7\right)\left(x-6\right)$$