Solve 2x^5-26x^4-96x^3

$$2 x ^ { 5 } - 26 x ^ { 4 } - 96 x ^ { 3 }$$

Factor

$$2\left(x-16\right)\left(x+3\right)x^{3}$$

Solution Steps

Factor out 2.

$$2\left(x^{5}-13x^{4}-48x^{3}\right)$$

Consider x^{5}-13x^{4}-48x^{3}. Factor out x^{3}.

$$x^{3}\left(x^{2}-13x-48\right)$$

Consider x^{2}-13x-48. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-48. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\left(-48\right)=-48$$

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 -48.

$$1,-48 2,-24 3,-16 4,-12 6,-8$$

Calculate the sum for each pair.

$$1-48=-47 2-24=-22 3-16=-13 4-12=-8 6-8=-2$$

The solution is the pair that gives sum -13.

$$a=-16 b=3$$

Rewrite x^{2}-13x-48 as \left(x^{2}-16x\right)+\left(3x-48\right).

$$\left(x^{2}-16x\right)+\left(3x-48\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-16\right)+3\left(x-16\right)$$

Factor out common term x-16 by using distributive property.

$$\left(x-16\right)\left(x+3\right)$$

Rewrite the complete factored expression.

$$2x^{3}\left(x-16\right)\left(x+3\right)$$

Evaluate

$$2\left(x-16\right)\left(x+3\right)x^{3}$$