Solve =4x^5-40x^3-12x^4+120x^2+36x-108

$$= 4 x ^ { 5 } - 40 x ^ { 3 } - 12 x ^ { 4 } + 120 x ^ { 2 } + 36 x - 108$$

Factor

$$4\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x-3\right)^{2}$$

Solution Steps

Factor out 4.

$$4\left(x^{5}-10x^{3}-3x^{4}+30x^{2}+9x-27\right)$$

Consider x^{5}-10x^{3}-3x^{4}+30x^{2}+9x-27. Find one factor of the form x^{k}+m, where x^{k} divides the monomial with the highest power x^{5} and m divides the constant factor -27. One such factor is x^{2}-9. Factor the polynomial by dividing it by this factor.

$$\left(x^{2}-9\right)\left(x^{3}-3x^{2}-x+3\right)$$

Consider x^{2}-9. Rewrite x^{2}-9 as x^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Consider x^{3}-3x^{2}-x+3. Do the grouping x^{3}-3x^{2}-x+3=\left(x^{3}-3x^{2}\right)+\left(-x+3\right), and factor out x^{2} in the first and -1 in the second group.

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

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x^{2}-1\right)$$

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$4\left(x-3\right)^{2}\left(x+3\right)\left(x-1\right)\left(x+1\right)$$

Evaluate

$$4\left(x+3\right)\left(x-3\right)^{2}\left(x^{2}-1\right)$$