Solve f(x)=-4x^6+4x^12+x^4+x

$$f ( x ) = - 4 x ^ { 6 } + 4 x ^ { 12 } + x ^ { 4 } + x$$

Factor

$$x\left(x+1\right)\left(x^{2}-x+1\right)\left(4x^{8}-4x^{5}+1\right)$$

Solution Steps

Factor out x.

$$x\left(-4x^{5}+4x^{11}+x^{3}+1\right)$$

Consider -4x^{5}+4x^{11}+x^{3}+1. Find one factor of the form kx^{m}+n, where kx^{m} divides the monomial with the highest power 4x^{11} and n divides the constant factor 1. One such factor is x^{3}+1. Factor the polynomial by dividing it by this factor.

$$\left(x^{3}+1\right)\left(4x^{8}-4x^{5}+1\right)$$

Consider x^{3}+1. Rewrite x^{3}+1 as x^{3}+1^{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).

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

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{2}-x+1,4x^{8}-4x^{5}+1.

$$x\left(x+1\right)\left(x^{2}-x+1\right)\left(4x^{8}-4x^{5}+1\right)$$

Evaluate

$$4x^{12}-4x^{6}+x^{4}+x$$