Solve (2x^2+5x-7)+(x-1)

$$(2 { x }^{ 2 } +5x-7)+(x-1)$$

Evaluate

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

Solution Steps

Combine 5x and x to get 6x.

$$2x^{2}+6x-7-1$$

Subtract 1 from -7 to get -8.

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

Factor

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

Solution Steps

Multiply and combine like terms.

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

Factor out 2.

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

Consider x^{2}+3x-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.

$$a+b=3 ab=1\left(-4\right)=-4$$

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 -4.

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

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

Rewrite x^{2}+3x-4 as \left(x^{2}-x\right)+\left(4x-4\right).

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

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

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

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

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

Rewrite the complete factored expression.

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