Solve f(x)=x^4-2x^3-8x^2

$$f ( x ) = x ^ { 4 } - 2 x ^ { 3 } - 8 x ^ { 2 }$$

Factor

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

Solution Steps

Factor out x^{2}.

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

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.

$$a+b=-2 ab=1\left(-8\right)=-8$$

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 -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.

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

Evaluate

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