Solve 9w^5+12w^4-21w^3

$$9 w ^ { 5 } + 12 w ^ { 4 } - 21 w ^ { 3 }$$

Factor

$$3\left(w-1\right)\left(3w+7\right)w^{3}$$

Solution Steps

Factor out 3.

$$3\left(3w^{5}+4w^{4}-7w^{3}\right)$$

Consider 3w^{5}+4w^{4}-7w^{3}. Factor out w^{3}.

$$w^{3}\left(3w^{2}+4w-7\right)$$

Consider 3w^{2}+4w-7. Factor the expression by grouping. First, the expression needs to be rewritten as 3w^{2}+aw+bw-7. To find a and b, set up a system to be solved.

$$a+b=4 ab=3\left(-7\right)=-21$$

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 -21.

$$-1,21 -3,7$$

Calculate the sum for each pair.

$$-1+21=20 -3+7=4$$

The solution is the pair that gives sum 4.

$$a=-3 b=7$$

Rewrite 3w^{2}+4w-7 as \left(3w^{2}-3w\right)+\left(7w-7\right).

$$\left(3w^{2}-3w\right)+\left(7w-7\right)$$

Factor out 3w in the first and 7 in the second group.

$$3w\left(w-1\right)+7\left(w-1\right)$$

Factor out common term w-1 by using distributive property.

$$\left(w-1\right)\left(3w+7\right)$$

Rewrite the complete factored expression.

$$3w^{3}\left(w-1\right)\left(3w+7\right)$$

Evaluate

$$3\left(w-1\right)\left(3w+7\right)w^{3}$$