Solve h(x)=x^4-2x^3-8x+16

$$h ( x ) = x ^ { 4 } - 2 x ^ { 3 } - 8 x + 16$$

Factor

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

Solution Steps

Do the grouping x^{4}-2x^{3}-8x+16=\left(x^{4}-2x^{3}\right)+\left(-8x+16\right), and factor out x^{3} in the first and -8 in the second group.

$$x^{3}\left(x-2\right)-8\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x^{3}-8\right)$$

Consider x^{3}-8. Rewrite x^{3}-8 as x^{3}-2^{3}. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right).

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

Rewrite the complete factored expression. Polynomial x^{2}+2x+4 is not factored since it does not have any rational roots.

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

Evaluate

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