Solve (9r-6)+(9r^2+8)

$$( 9 r - 6 ) + ( 9 r ^ { 2 } + 8 )$$

Factor

$$\left(3r+1\right)\left(3r+2\right)$$

Solution Steps

Multiply and combine like terms.

$$9r^{2}+9r+2$$

Factor the expression by grouping. First, the expression needs to be rewritten as 9r^{2}+ar+br+2. To find a and b, set up a system to be solved.

$$a+b=9 ab=9\times 2=18$$

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 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 9.

$$a=3 b=6$$

Rewrite 9r^{2}+9r+2 as \left(9r^{2}+3r\right)+\left(6r+2\right).

$$\left(9r^{2}+3r\right)+\left(6r+2\right)$$

Factor out 3r in the first and 2 in the second group.

$$3r\left(3r+1\right)+2\left(3r+1\right)$$

Factor out common term 3r+1 by using distributive property.

$$\left(3r+1\right)\left(3r+2\right)$$

Evaluate

$$\left(3r+1\right)\left(3r+2\right)$$

Solution Steps

Add -6 and 8 to get 2.

$$9r+2+9r^{2}$$