Solve 2x^6+32x^2-16x^4

$$2 x ^ { 6 } + 32 x ^ { 2 } - 16 x ^ { 4 }$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(x^{6}+16x^{2}-8x^{4}\right)$$

Consider x^{6}+16x^{2}-8x^{4}. Factor out x^{2}.

$$x^{2}\left(x^{4}+16-8x^{2}\right)$$

Consider x^{4}+16-8x^{2}. Find one factor of the form x^{k}+m, where x^{k} divides the monomial with the highest power x^{4} and m divides the constant factor 16. One such factor is x^{2}-4. Factor the polynomial by dividing it by this factor.

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

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.

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

Evaluate

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