Solve (3r^2+5r-6)+(2r-5r^2)

$$( 3 r ^ { 2 } + 5 r - 6 ) + ( 2 r - 5 r ^ { 2 } )$$

Evaluate

$$\left(2-r\right)\left(2r-3\right)$$

Solution Steps

Combine 5r and 2r to get 7r.

$$3r^{2}+7r-6-5r^{2}$$

Combine 3r^{2} and -5r^{2} to get -2r^{2}.

$$-2r^{2}+7r-6$$

Factor

$$\left(2-r\right)\left(2r-3\right)$$

Solution Steps

Multiply and combine like terms.

$$-2r^{2}+7r-6$$

Factor the expression by grouping. First, the expression needs to be rewritten as -2r^{2}+ar+br-6. To find a and b, set up a system to be solved.

$$a+b=7 ab=-2\left(-6\right)=12$$

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 12.

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

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=4 b=3$$

Rewrite -2r^{2}+7r-6 as \left(-2r^{2}+4r\right)+\left(3r-6\right).

$$\left(-2r^{2}+4r\right)+\left(3r-6\right)$$

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

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

Factor out common term -r+2 by using distributive property.

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