Solve f(x)=1/4x^4-x^3-x^2+4x

$$f ( x ) = \frac { 1 } { 4 } x ^ { 4 } - x ^ { 3 } - x ^ { 2 } + 4 x$$

Factor

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

Solution Steps

Factor out \frac{1}{4}.

$$\frac{x^{4}-4x^{3}-4x^{2}+16x}{4}$$

Consider x^{4}-4x^{3}-4x^{2}+16x. Factor out x.

$$x\left(x^{3}-4x^{2}-4x+16\right)$$

Consider x^{3}-4x^{2}-4x+16. Do the grouping x^{3}-4x^{2}-4x+16=\left(x^{3}-4x^{2}\right)+\left(-4x+16\right), and factor out x^{2} in the first and -4 in the second group.

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

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

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

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.

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

Evaluate

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