Solve x^3-3x^2-28x

$$x ^ { 3 } - 3 x ^ { 2 } - 28 x$$

Factor

$$x\left(x-7\right)\left(x+4\right)$$

Solution Steps

Factor out x.

$$x\left(x^{2}-3x-28\right)$$

Consider x^{2}-3x-28. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-28\right)=-28$$

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 -28.

$$1,-28 2,-14 4,-7$$

Calculate the sum for each pair.

$$1-28=-27 2-14=-12 4-7=-3$$

The solution is the pair that gives sum -3.

$$a=-7 b=4$$

Rewrite x^{2}-3x-28 as \left(x^{2}-7x\right)+\left(4x-28\right).

$$\left(x^{2}-7x\right)+\left(4x-28\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x-7\right)+4\left(x-7\right)$$

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

$$\left(x-7\right)\left(x+4\right)$$

Rewrite the complete factored expression.

$$x\left(x-7\right)\left(x+4\right)$$

Evaluate

$$x\left(x-7\right)\left(x+4\right)$$