Solve (x^2-8x-2)+(4x-3)

$$( { x }^{ 2 } -8x-2)+(4x-3)$$

Evaluate

$$\left(x-5\right)\left(x+1\right)$$

Solution Steps

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

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

Subtract 3 from -2 to get -5.

$$x^{2}-4x-5$$

Factor

$$\left(x-5\right)\left(x+1\right)$$

Solution Steps

Multiply and combine like terms.

$$x^{2}-4x-5$$

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-5\right)=-5$$

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. The only such pair is the system solution.

$$a=-5 b=1$$

Rewrite x^{2}-4x-5 as \left(x^{2}-5x\right)+\left(x-5\right).

$$\left(x^{2}-5x\right)+\left(x-5\right)$$

Factor out x in x^{2}-5x.

$$x\left(x-5\right)+x-5$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+1\right)$$