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

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

Evaluate

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

Solution Steps

Combine 3x^{2} and 2x^{2} to get 5x^{2}.

$$5x^{2}-6x+4+4x-7$$

Combine -6x and 4x to get -2x.

$$5x^{2}-2x+4-7$$

Subtract 7 from 4 to get -3.

$$5x^{2}-2x-3$$

Factor

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

Solution Steps

Multiply and combine like terms.

$$5x^{2}-2x-3$$

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

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

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

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

Factor out 5x in the first and 3 in the second group.

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

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

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