Solve =-x^5-6x^4-9x^3+4x^2+12x

$$= - x ^ { 5 } - 6 x ^ { 4 } - 9 x ^ { 3 } + 4 x ^ { 2 } + 12 x$$

Factor

$$x\left(1-x\right)\left(x+3\right)\left(x+2\right)^{2}$$

Solution Steps

Factor out x.

$$x\left(-x^{4}-6x^{3}-9x^{2}+4x+12\right)$$

Consider -x^{4}-6x^{3}-9x^{2}+4x+12. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 12 and q divides the leading coefficient -1. One such root is -3. Factor the polynomial by dividing it by x+3.

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

Consider -x^{3}-3x^{2}+4. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 4 and q divides the leading coefficient -1. One such root is -2. Factor the polynomial by dividing it by x+2.

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

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

$$a+b=-1 ab=-2=-2$$

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. The only such pair is the system solution.

$$a=1 b=-2$$

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

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

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

$$x\left(-x+1\right)+2\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+2\right)$$

Rewrite the complete factored expression.

$$x\left(x+3\right)\left(x+2\right)^{2}\left(-x+1\right)$$

Evaluate

$$x\left(1-x\right)\left(x+3\right)\left(x+2\right)^{2}$$