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

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

Evaluate

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

Solution Steps

Combine 4x^{2} and 2x^{2} to get 6x^{2}.

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

Combine 7x and 3x to get 10x.

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

Factor

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

Solution Steps

Multiply and combine like terms.

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

Factor out 2.

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

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

$$a+b=5 ab=3\left(-2\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 5.

$$a=-1 b=6$$

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

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

Factor out x in the first and 2 in the second group.

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

Factor out common term 3x-1 by using distributive property.

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

Rewrite the complete factored expression.

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