Solve 243x^5-108x^3+72x^2-32

$$243 x ^ { 5 } - 108 x ^ { 3 } + 72 x ^ { 2 } - 32$$

Factor

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

Solution Steps

Do the grouping 243x^{5}-108x^{3}+72x^{2}-32=\left(243x^{5}-108x^{3}\right)+\left(72x^{2}-32\right), and factor out 27x^{3} in the first and 8 in the second group.

$$27x^{3}\left(9x^{2}-4\right)+8\left(9x^{2}-4\right)$$

Factor out common term 9x^{2}-4 by using distributive property.

$$\left(9x^{2}-4\right)\left(27x^{3}+8\right)$$

Consider 9x^{2}-4. Rewrite 9x^{2}-4 as \left(3x\right)^{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).

$$\left(3x-2\right)\left(3x+2\right)$$

Consider 27x^{3}+8. Rewrite 27x^{3}+8 as \left(3x\right)^{3}+2^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

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

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

Evaluate

$$243x^{5}-108x^{3}+72x^{2}-32$$