Solve (a^4-8a^2b^2+16b^4)

$$( a ^ { 4 } - 8 a ^ { 2 } b ^ { 2 } + 16 b ^ { 4 } )$$

Factor

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

Solution Steps

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

$$a^{4}-8b^{2}a^{2}+16b^{4}$$

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

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

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

$$\left(a-2b\right)\left(a+2b\right)$$

Rewrite the complete factored expression.

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

Evaluate

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