Solve 3x^2-2x+6x+2+6x^2+4x-3

$$3 { x }^{ 2 } -2x+6x+2+6 { x }^{ 2 } +4x-3$$

Evaluate

$$\left(9x-1\right)\left(x+1\right)$$

Solution Steps

Combine -2x and 6x to get 4x.

$$3x^{2}+4x+2+6x^{2}+4x-3$$

Combine 3x^{2} and 6x^{2} to get 9x^{2}.

$$9x^{2}+4x+2+4x-3$$

Combine 4x and 4x to get 8x.

$$9x^{2}+8x+2-3$$

Subtract 3 from 2 to get -1.

$$9x^{2}+8x-1$$

Factor

$$\left(9x-1\right)\left(x+1\right)$$

Solution Steps

Multiply and combine like terms.

$$9x^{2}+8x-1$$

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=8 ab=9\left(-1\right)=-9$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -9.

$$-1,9 -3,3$$

Calculate the sum for each pair.

$$-1+9=8 -3+3=0$$

The solution is the pair that gives sum 8.

$$a=-1 b=9$$

Rewrite 9x^{2}+8x-1 as \left(9x^{2}-x\right)+\left(9x-1\right).

$$\left(9x^{2}-x\right)+\left(9x-1\right)$$

Factor out x in 9x^{2}-x.

$$x\left(9x-1\right)+9x-1$$

Factor out common term 9x-1 by using distributive property.

$$\left(9x-1\right)\left(x+1\right)$$