Solve 2x+4+x^2+2x-8x

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

Evaluate

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

Solution Steps

Combine 2x and 2x to get 4x.

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

Combine 4x and -8x to get -4x.

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

Factor

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

Solution Steps

Multiply and combine like terms.

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

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

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

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

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

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

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

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

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

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

Rewrite as a binomial square.

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