Solve u^2+3u+8u+24

$$u ^ { 2 } + 3 u + 8 u + 24$$

Factor

$$\left(u+3\right)\left(u+8\right)$$

Solution Steps

Multiply and combine like terms.

$$u^{2}+11u+24$$

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

$$a+b=11 ab=1\times 24=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 11.

$$a=3 b=8$$

Rewrite u^{2}+11u+24 as \left(u^{2}+3u\right)+\left(8u+24\right).

$$\left(u^{2}+3u\right)+\left(8u+24\right)$$

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

$$u\left(u+3\right)+8\left(u+3\right)$$

Factor out common term u+3 by using distributive property.

$$\left(u+3\right)\left(u+8\right)$$

Evaluate

$$\left(u+3\right)\left(u+8\right)$$

Solution Steps

Combine 3u and 8u to get 11u.

$$u^{2}+11u+24$$