Solve 20k^3+12k^2-5k-3=

$$20 k ^ { 3 } + 12 k ^ { 2 } - 5 k - 3 =$$

Factor

$$\left(2k-1\right)\left(2k+1\right)\left(5k+3\right)$$

Solution Steps

Do the grouping 20k^{3}+12k^{2}-5k-3=\left(20k^{3}+12k^{2}\right)+\left(-5k-3\right), and factor out 4k^{2} in the first and -1 in the second group.

$$4k^{2}\left(5k+3\right)-\left(5k+3\right)$$

Factor out common term 5k+3 by using distributive property.

$$\left(5k+3\right)\left(4k^{2}-1\right)$$

Consider 4k^{2}-1. Rewrite 4k^{2}-1 as \left(2k\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2k-1\right)\left(2k+1\right)$$

Rewrite the complete factored expression.

$$\left(2k-1\right)\left(2k+1\right)\left(5k+3\right)$$

Evaluate

$$\left(5k+3\right)\left(4k^{2}-1\right)$$