Solve (9x^2-12x)+(6x-8)

$$(9 { x }^{ 2 } -12x)+(6x-8)$$

Factor

$$\left(3x-4\right)\left(3x+2\right)$$

Solution Steps

Multiply and combine like terms.

$$9x^{2}-6x-8$$

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=9\left(-8\right)=-72$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -6.

$$a=-12 b=6$$

Rewrite 9x^{2}-6x-8 as \left(9x^{2}-12x\right)+\left(6x-8\right).

$$\left(9x^{2}-12x\right)+\left(6x-8\right)$$

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

$$3x\left(3x-4\right)+2\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(3x+2\right)$$

Evaluate

$$\left(3x-4\right)\left(3x+2\right)$$

Solution Steps

Combine -12x and 6x to get -6x.

$$9x^{2}-6x-8$$