Solve 16x^4-8x^2y^2+y^4

$$16 x ^ { 4 } - 8 x ^ { 2 } y ^ { 2 } + y ^ { 4 }$$

Factor

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

Solution Steps

Consider 16x^{4}-8x^{2}y^{2}+y^{4} as a polynomial over variable x.

$$16x^{4}-8y^{2}x^{2}+y^{4}$$

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

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

Consider 4x^{2}-y^{2}. Rewrite 4x^{2}-y^{2} as \left(2x\right)^{2}-y^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Rewrite the complete factored expression.

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

Evaluate

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