Solve {l}{a^{4}+4a^{3}+6a^{2}}{+4a+1-4a^3}{-8a^2+4a}{+2a^2-1}{+a^4}

$$\left. \begin{array} { l } { a ^ { 4 } + 4 a ^ { 3 } + 6 a ^ { 2 } } \\ { + 4 a + 1 - 4 a ^ { 3 } } \\ { - 8 a ^ { 2 } + 4 a } \\ { + 2 a ^ { 2 } - 1 } \\ { + a ^ { 4 } } \end{array} \right.$$

Evaluate

$$2a\left(a^{3}+4\right)$$

Solution Steps

Combine 4a^{3} and -4a^{3} to get 0.

$$a^{4}+6a^{2}+4a+1-8a^{2}+4a+2a^{2}-1+a^{4}$$

Combine 6a^{2} and -8a^{2} to get -2a^{2}.

$$a^{4}-2a^{2}+4a+1+4a+2a^{2}-1+a^{4}$$

Combine 4a and 4a to get 8a.

$$a^{4}-2a^{2}+8a+1+2a^{2}-1+a^{4}$$

Combine -2a^{2} and 2a^{2} to get 0.

$$a^{4}+8a+1-1+a^{4}$$

Subtract 1 from 1 to get 0.

$$a^{4}+8a+a^{4}$$

Combine a^{4} and a^{4} to get 2a^{4}.

$$2a^{4}+8a$$

Factor

$$2a\left(a^{3}+4\right)$$

Solution Steps

Multiply and combine like terms.

$$2a^{4}+8a$$

Factor out 2.

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

Consider a^{4}+4a. Factor out a.

$$a\left(a^{3}+4\right)$$

Rewrite the complete factored expression. Polynomial a^{3}+4 is not factored since it does not have any rational roots.

$$2a\left(a^{3}+4\right)$$