Solve (c^2+14c+45)+(c+5)

$$( c ^ { 2 } + 14 c + 45 ) + ( c + 5 )$$

Evaluate

$$\left(c+5\right)\left(c+10\right)$$

Solution Steps

Combine 14c and c to get 15c.

$$c^{2}+15c+45+5$$

Add 45 and 5 to get 50.

$$c^{2}+15c+50$$

Factor

$$\left(c+5\right)\left(c+10\right)$$

Solution Steps

Multiply and combine like terms.

$$c^{2}+15c+50$$

Factor the expression by grouping. First, the expression needs to be rewritten as c^{2}+ac+bc+50. To find a and b, set up a system to be solved.

$$a+b=15 ab=1\times 50=50$$

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 50.

$$1,50 2,25 5,10$$

Calculate the sum for each pair.

$$1+50=51 2+25=27 5+10=15$$

The solution is the pair that gives sum 15.

$$a=5 b=10$$

Rewrite c^{2}+15c+50 as \left(c^{2}+5c\right)+\left(10c+50\right).

$$\left(c^{2}+5c\right)+\left(10c+50\right)$$

Factor out c in the first and 10 in the second group.

$$c\left(c+5\right)+10\left(c+5\right)$$

Factor out common term c+5 by using distributive property.

$$\left(c+5\right)\left(c+10\right)$$