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

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

Factor

$$4\left(3x-1\right)\left(2x+1\right)$$

Solution Steps

Factor out 4.

$$4\left(6x^{2}+x-1\right)$$

Consider 6x^{2}+x-1. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=1 ab=6\left(-1\right)=-6$$

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 1.

$$a=-2 b=3$$

Rewrite 6x^{2}+x-1 as \left(6x^{2}-2x\right)+\left(3x-1\right).

$$\left(6x^{2}-2x\right)+\left(3x-1\right)$$

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

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

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

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

Rewrite the complete factored expression.

$$4\left(3x-1\right)\left(2x+1\right)$$

Evaluate

$$4\left(6x^{2}+x-1\right)$$

Solution Steps

Multiply 8 and 3 to get 24.

$$24x^{2}+4x-4$$