Solve 8x^5-32x^3-27x^2+108

$$8 x ^ { 5 } - 32 x ^ { 3 } - 27 x ^ { 2 } + 108$$

Factor

$$\left(x-2\right)\left(2x-3\right)\left(x+2\right)\left(4x^{2}+6x+9\right)$$

Solution Steps

Do the grouping 8x^{5}-32x^{3}-27x^{2}+108=\left(8x^{5}-32x^{3}\right)+\left(-27x^{2}+108\right), and factor out 8x^{3} in the first and -27 in the second group.

$$8x^{3}\left(x^{2}-4\right)-27\left(x^{2}-4\right)$$

Factor out common term x^{2}-4 by using distributive property.

$$\left(x^{2}-4\right)\left(8x^{3}-27\right)$$

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Consider 8x^{3}-27. Rewrite 8x^{3}-27 as \left(2x\right)^{3}-3^{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).

$$\left(2x-3\right)\left(4x^{2}+6x+9\right)$$

Rewrite the complete factored expression. Polynomial 4x^{2}+6x+9 is not factored since it does not have any rational roots.

$$\left(2x-3\right)\left(x-2\right)\left(x+2\right)\left(4x^{2}+6x+9\right)$$

Evaluate

$$8x^{5}-32x^{3}-27x^{2}+108$$