Solve 6k^4+66k^3+144k^2

$$6 k ^ { 4 } + 66 k ^ { 3 } + 144 k ^ { 2 }$$

Factor

$$6\left(k+3\right)\left(k+8\right)k^{2}$$

Solution Steps

Factor out 6.

$$6\left(k^{4}+11k^{3}+24k^{2}\right)$$

Consider k^{4}+11k^{3}+24k^{2}. Factor out k^{2}.

$$k^{2}\left(k^{2}+11k+24\right)$$

Consider k^{2}+11k+24. Factor the expression by grouping. First, the expression needs to be rewritten as k^{2}+ak+bk+24. To find a and b, set up a system to be solved.

$$a+b=11 ab=1\times 24=24$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 11.

$$a=3 b=8$$

Rewrite k^{2}+11k+24 as \left(k^{2}+3k\right)+\left(8k+24\right).

$$\left(k^{2}+3k\right)+\left(8k+24\right)$$

Factor out k in the first and 8 in the second group.

$$k\left(k+3\right)+8\left(k+3\right)$$

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

$$\left(k+3\right)\left(k+8\right)$$

Rewrite the complete factored expression.

$$6k^{2}\left(k+3\right)\left(k+8\right)$$

Evaluate

$$6\left(k+3\right)\left(k+8\right)k^{2}$$