Solve ax^4-8ax^2+16a-x^4+8x^2-16=

$$a x ^ { 4 } - 8 a x ^ { 2 } + 16 a - x ^ { 4 } + 8 x ^ { 2 } - 16 =$$

Factor

$$\left(a-1\right)\left(x^{2}-4\right)^{2}$$

Solution Steps

Do the grouping ax^{4}-8ax^{2}+16a-x^{4}+8x^{2}-16=\left(ax^{4}-8ax^{2}+16a\right)+\left(-x^{4}+8x^{2}-16\right), and factor out a in the first and -1 in the second group.

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

Factor out common term x^{4}-8x^{2}+16 by using distributive property.

$$\left(x^{4}-8x^{2}+16\right)\left(a-1\right)$$

Consider x^{4}-8x^{2}+16. 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: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Rewrite the complete factored expression.

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

Evaluate

$$\left(a-1\right)\left(x^{2}-4\right)^{2}$$