Solve x^3-x^2-42x=

$$x ^ { 3 } - x ^ { 2 } - 42 x =$$

Factor

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

Solution Steps

Factor out x.

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

Consider x^{2}-x-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.

$$a+b=-1 ab=1\left(-42\right)=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -42.

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

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -1.

$$a=-7 b=6$$

Rewrite x^{2}-x-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

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