Solve 10w^4+32w^3+6w^2

$$10 w ^ { 4 } + 32 w ^ { 3 } + 6 w ^ { 2 }$$

Factor

$$2\left(w+3\right)\left(5w+1\right)w^{2}$$

Solution Steps

Factor out 2.

$$2\left(5w^{4}+16w^{3}+3w^{2}\right)$$

Consider 5w^{4}+16w^{3}+3w^{2}. Factor out w^{2}.

$$w^{2}\left(5w^{2}+16w+3\right)$$

Consider 5w^{2}+16w+3. Factor the expression by grouping. First, the expression needs to be rewritten as 5w^{2}+aw+bw+3. To find a and b, set up a system to be solved.

$$a+b=16 ab=5\times 3=15$$

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 15.

$$1,15 3,5$$

Calculate the sum for each pair.

$$1+15=16 3+5=8$$

The solution is the pair that gives sum 16.

$$a=1 b=15$$

Rewrite 5w^{2}+16w+3 as \left(5w^{2}+w\right)+\left(15w+3\right).

$$\left(5w^{2}+w\right)+\left(15w+3\right)$$

Factor out w in the first and 3 in the second group.

$$w\left(5w+1\right)+3\left(5w+1\right)$$

Factor out common term 5w+1 by using distributive property.

$$\left(5w+1\right)\left(w+3\right)$$

Rewrite the complete factored expression.

$$2w^{2}\left(5w+1\right)\left(w+3\right)$$

Evaluate

$$2\left(w+3\right)\left(5w+1\right)w^{2}$$