Solve 3w^4+24w^3+45w^2=

$$3 w ^ { 4 } + 24 w ^ { 3 } + 45 w ^ { 2 } =$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(w^{4}+8w^{3}+15w^{2}\right)$$

Consider w^{4}+8w^{3}+15w^{2}. Factor out w^{2}.

$$w^{2}\left(w^{2}+8w+15\right)$$

Consider w^{2}+8w+15. Factor the expression by grouping. First, the expression needs to be rewritten as w^{2}+aw+bw+15. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\times 15=15$$

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 8.

$$a=3 b=5$$

Rewrite w^{2}+8w+15 as \left(w^{2}+3w\right)+\left(5w+15\right).

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

Factor out w in the first and 5 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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