Solve x^5-4x^3+3/2x^2-6

$${ x }^{ 5 } -4 { x }^{ 3 } + \frac{ 3 }{ 2 } { x }^{ 2 } -6$$

Factor

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

Solution Steps

Factor out \frac{1}{2}.

$$\frac{2x^{5}-8x^{3}+3x^{2}-12}{2}$$

Consider 2x^{5}-8x^{3}+3x^{2}-12. Do the grouping 2x^{5}-8x^{3}+3x^{2}-12=\left(2x^{5}-8x^{3}\right)+\left(3x^{2}-12\right), and factor out 2x^{3} in the first and 3 in the second group.

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

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

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

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).

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

Rewrite the complete factored expression. Polynomial 2x^{3}+3 is not factored since it does not have any rational roots.

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

Evaluate

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