Solve (2x^2-6x-1)+(8x^2+5x-2)

$$( 2 x ^ { 2 } - 6 x - 1 ) + ( 8 x ^ { 2 } + 5 x - 2 )$$

Evaluate

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

Solution Steps

Combine 2x^{2} and 8x^{2} to get 10x^{2}.

$$10x^{2}-6x-1+5x-2$$

Combine -6x and 5x to get -x.

$$10x^{2}-x-1-2$$

Subtract 2 from -1 to get -3.

$$10x^{2}-x-3$$

Factor

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

Solution Steps

Multiply and combine like terms.

$$10x^{2}-x-3$$

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-1 ab=10\left(-3\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=-6 b=5$$

Rewrite 10x^{2}-x-3 as \left(10x^{2}-6x\right)+\left(5x-3\right).

$$\left(10x^{2}-6x\right)+\left(5x-3\right)$$

Factor out 2x in 10x^{2}-6x.

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

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

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