Solve 2x^6-32x^5-72x^4

$$2 x ^ { 6 } - 32 x ^ { 5 } - 72 x ^ { 4 }$$

Factor

$$2\left(x-18\right)\left(x+2\right)x^{4}$$

Solution Steps

Factor out 2.

$$2\left(x^{6}-16x^{5}-36x^{4}\right)$$

Consider x^{6}-16x^{5}-36x^{4}. Factor out x^{4}.

$$x^{4}\left(x^{2}-16x-36\right)$$

Consider x^{2}-16x-36. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-36. To find a and b, set up a system to be solved.

$$a+b=-16 ab=1\left(-36\right)=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -16.

$$a=-18 b=2$$

Rewrite x^{2}-16x-36 as \left(x^{2}-18x\right)+\left(2x-36\right).

$$\left(x^{2}-18x\right)+\left(2x-36\right)$$

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

$$x\left(x-18\right)+2\left(x-18\right)$$

Factor out common term x-18 by using distributive property.

$$\left(x-18\right)\left(x+2\right)$$

Rewrite the complete factored expression.

$$2x^{4}\left(x-18\right)\left(x+2\right)$$

Evaluate

$$2\left(x-18\right)\left(x+2\right)x^{4}$$