Solve 5m^2+40m+3m+24

$$5 m ^ { 2 } + 40 m + 3 m + 24$$

Factor

$$\left(m+8\right)\left(5m+3\right)$$

Solution Steps

Multiply and combine like terms.

$$5m^{2}+43m+24$$

Factor the expression by grouping. First, the expression needs to be rewritten as 5m^{2}+am+bm+24. To find a and b, set up a system to be solved.

$$a+b=43 ab=5\times 24=120$$

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 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 43.

$$a=3 b=40$$

Rewrite 5m^{2}+43m+24 as \left(5m^{2}+3m\right)+\left(40m+24\right).

$$\left(5m^{2}+3m\right)+\left(40m+24\right)$$

Factor out m in the first and 8 in the second group.

$$m\left(5m+3\right)+8\left(5m+3\right)$$

Factor out common term 5m+3 by using distributive property.

$$\left(5m+3\right)\left(m+8\right)$$

Evaluate

$$\left(m+8\right)\left(5m+3\right)$$

Solution Steps

Combine 40m and 3m to get 43m.

$$5m^{2}+43m+24$$