Solve 2x^2+6x+4-12-x^2-4x

$$2 { x }^{ 2 } +6x+4-12- { x }^{ 2 } -4x$$

Evaluate

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

Solution Steps

Subtract 12 from 4 to get -8.

$$2x^{2}+6x-8-x^{2}-4x$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}+6x-8-4x$$

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

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

Factor

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

Solution Steps

Multiply and combine like terms.

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

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

$$a+b=2 ab=1\left(-8\right)=-8$$

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 -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=-2 b=4$$

Rewrite x^{2}+2x-8 as \left(x^{2}-2x\right)+\left(4x-8\right).

$$\left(x^{2}-2x\right)+\left(4x-8\right)$$

Factor out x in the first and 4 in the second group.

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

Factor out common term x-2 by using distributive property.

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