Solve 4k^3+4k^2+6k+6

$$4 k ^ { 3 } + 4 k ^ { 2 } + 6 k + 6$$

Factor

$$2\left(k+1\right)\left(2k^{2}+3\right)$$

Solution Steps

Factor out 2.

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

Consider 2k^{3}+2k^{2}+3k+3. Do the grouping 2k^{3}+2k^{2}+3k+3=\left(2k^{3}+2k^{2}\right)+\left(3k+3\right), and factor out 2k^{2} in the first and 3 in the second group.

$$2k^{2}\left(k+1\right)+3\left(k+1\right)$$

Factor out common term k+1 by using distributive property.

$$\left(k+1\right)\left(2k^{2}+3\right)$$

Rewrite the complete factored expression. Polynomial 2k^{2}+3 is not factored since it does not have any rational roots.

$$2\left(k+1\right)\left(2k^{2}+3\right)$$

Evaluate

$$2\left(k+1\right)\left(2k^{2}+3\right)$$