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

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

Factor

$$\left(x-4\right)\left(2x+1\right)$$

Solution Steps

Multiply and combine like terms.

$$2x^{2}-7x-4$$

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

$$a+b=-7 ab=2\left(-4\right)=-8$$

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 -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 -7.

$$a=-8 b=1$$

Rewrite 2x^{2}-7x-4 as \left(2x^{2}-8x\right)+\left(x-4\right).

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

Factor out 2x in 2x^{2}-8x.

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

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

$$\left(x-4\right)\left(2x+1\right)$$

Evaluate

$$\left(x-4\right)\left(2x+1\right)$$

Solution Steps

Combine -8x and x to get -7x.

$$2x^{2}-7x-4$$