Solve 3x^5-18x^4-48x^3

$$3 { x }^{ 5 } -18 { x }^{ 4 } -48 { x }^{ 3 }$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(x^{5}-6x^{4}-16x^{3}\right)$$

Consider x^{5}-6x^{4}-16x^{3}. Factor out x^{3}.

$$x^{3}\left(x^{2}-6x-16\right)$$

Consider x^{2}-6x-16. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-16\right)=-16$$

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 -16.

$$1,-16 2,-8 4,-4$$

Calculate the sum for each pair.

$$1-16=-15 2-8=-6 4-4=0$$

The solution is the pair that gives sum -6.

$$a=-8 b=2$$

Rewrite x^{2}-6x-16 as \left(x^{2}-8x\right)+\left(2x-16\right).

$$\left(x^{2}-8x\right)+\left(2x-16\right)$$

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

$$x\left(x-8\right)+2\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x+2\right)$$

Rewrite the complete factored expression.

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

Evaluate

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