Solve (x^2-7x-6)+(5x^2+3x-4)=

$$( x ^ { 2 } - 7 x - 6 ) + ( 5 x ^ { 2 } + 3 x - 4 ) =$$

Evaluate

$$2\left(3x-5\right)\left(x+1\right)$$

Solution Steps

Combine x^{2} and 5x^{2} to get 6x^{2}.

$$6x^{2}-7x-6+3x-4$$

Combine -7x and 3x to get -4x.

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

Subtract 4 from -6 to get -10.

$$6x^{2}-4x-10$$

Factor

$$2\left(3x-5\right)\left(x+1\right)$$

Solution Steps

Multiply and combine like terms.

$$6x^{2}-4x-10$$

Factor out 2.

$$2\left(3x^{2}-2x-5\right)$$

Consider 3x^{2}-2x-5. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-2 ab=3\left(-5\right)=-15$$

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 -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite 3x^{2}-2x-5 as \left(3x^{2}-5x\right)+\left(3x-5\right).

$$\left(3x^{2}-5x\right)+\left(3x-5\right)$$

Factor out x in 3x^{2}-5x.

$$x\left(3x-5\right)+3x-5$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$2\left(3x-5\right)\left(x+1\right)$$