Solve x^3-7x^2+8x+16

$${ x }^{ 3 } -7 { x }^{ 2 } +8x+16$$

Factor

$$\left(x+1\right)\left(x-4\right)^{2}$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 16 and q divides the leading coefficient 1. One such root is 4. Factor the polynomial by dividing it by x-4.

$$\left(x-4\right)\left(x^{2}-3x-4\right)$$

Consider x^{2}-3x-4. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-4\right)=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

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

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

Factor out x in x^{2}-4x.

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

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$\left(x+1\right)\left(x-4\right)^{2}$$

Evaluate

$$\left(x+1\right)\left(x-4\right)^{2}$$