Solve -x^3+6x^2-32

$$- x ^ { 3 } + 6 x ^ { 2 } - 32$$

Factor

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

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 -32 and q divides the leading coefficient -1. One such root is 4. Factor the polynomial by dividing it by x-4.

$$\left(x-4\right)\left(-x^{2}+2x+8\right)$$

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

$$a+b=2 ab=-8=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=4 b=-2$$

Rewrite -x^{2}+2x+8 as \left(-x^{2}+4x\right)+\left(-2x+8\right).

$$\left(-x^{2}+4x\right)+\left(-2x+8\right)$$

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

$$-x\left(x-4\right)-2\left(x-4\right)$$

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

$$\left(x-4\right)\left(-x-2\right)$$

Rewrite the complete factored expression.

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

Evaluate

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