Solve 21w^2+63w+15w+45

$$21 w ^ { 2 } + 63 w + 15 w + 45$$

Factor

$$3\left(w+3\right)\left(7w+5\right)$$

Solution Steps

Factor out 3.

$$3\left(7w^{2}+21w+5w+15\right)$$

Consider 7w^{2}+21w+5w+15. Multiply and combine like terms.

$$7w^{2}+26w+15$$

Consider 7w^{2}+26w+15. Factor the expression by grouping. First, the expression needs to be rewritten as 7w^{2}+aw+bw+15. To find a and b, set up a system to be solved.

$$a+b=26 ab=7\times 15=105$$

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 105.

$$1,105 3,35 5,21 7,15$$

Calculate the sum for each pair.

$$1+105=106 3+35=38 5+21=26 7+15=22$$

The solution is the pair that gives sum 26.

$$a=5 b=21$$

Rewrite 7w^{2}+26w+15 as \left(7w^{2}+5w\right)+\left(21w+15\right).

$$\left(7w^{2}+5w\right)+\left(21w+15\right)$$

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

$$w\left(7w+5\right)+3\left(7w+5\right)$$

Factor out common term 7w+5 by using distributive property.

$$\left(7w+5\right)\left(w+3\right)$$

Rewrite the complete factored expression.

$$3\left(7w+5\right)\left(w+3\right)$$

Evaluate

$$3\left(w+3\right)\left(7w+5\right)$$

Solution Steps

Combine 63w and 15w to get 78w.

$$21w^{2}+78w+45$$