Solve 12w^4+40w^3+32w^2

$$12 w ^ { 4 } + 40 w ^ { 3 } + 32 w ^ { 2 }$$

Factor

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

Solution Steps

Factor out 4.

$$4\left(3w^{4}+10w^{3}+8w^{2}\right)$$

Consider 3w^{4}+10w^{3}+8w^{2}. Factor out w^{2}.

$$w^{2}\left(3w^{2}+10w+8\right)$$

Consider 3w^{2}+10w+8. Factor the expression by grouping. First, the expression needs to be rewritten as 3w^{2}+aw+bw+8. To find a and b, set up a system to be solved.

$$a+b=10 ab=3\times 8=24$$

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 10.

$$a=4 b=6$$

Rewrite 3w^{2}+10w+8 as \left(3w^{2}+4w\right)+\left(6w+8\right).

$$\left(3w^{2}+4w\right)+\left(6w+8\right)$$

Factor out w in the first and 2 in the second group.

$$w\left(3w+4\right)+2\left(3w+4\right)$$

Factor out common term 3w+4 by using distributive property.

$$\left(3w+4\right)\left(w+2\right)$$

Rewrite the complete factored expression.

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

Evaluate

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