Solve f(x)=1/2x^3-2x^2-1/2x+2

$$f ( x ) = \frac { 1 } { 2 } x ^ { 3 } - 2 x ^ { 2 } - \frac { 1 } { 2 } x + 2$$

Factor

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

Solution Steps

Factor out \frac{1}{2}.

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

Consider x^{3}-4x^{2}-x+4. Do the grouping x^{3}-4x^{2}-x+4=\left(x^{3}-4x^{2}\right)+\left(-x+4\right), and factor out x^{2} in the first and -1 in the second group.

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

Factor out common term x-4 by using distributive property.

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

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.

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

Evaluate

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