Solve 10w^5+15w^4-45w^3

$$10 w ^ { 5 } + 15 w ^ { 4 } - 45 w ^ { 3 }$$

Factor

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

Solution Steps

Factor out 5.

$$5\left(2w^{5}+3w^{4}-9w^{3}\right)$$

Consider 2w^{5}+3w^{4}-9w^{3}. Factor out w^{3}.

$$w^{3}\left(2w^{2}+3w-9\right)$$

Consider 2w^{2}+3w-9. Factor the expression by grouping. First, the expression needs to be rewritten as 2w^{2}+aw+bw-9. To find a and b, set up a system to be solved.

$$a+b=3 ab=2\left(-9\right)=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 3.

$$a=-3 b=6$$

Rewrite 2w^{2}+3w-9 as \left(2w^{2}-3w\right)+\left(6w-9\right).

$$\left(2w^{2}-3w\right)+\left(6w-9\right)$$

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

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

Factor out common term 2w-3 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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