Solve -8x-8+6x^2

$$- 8 x - 8 + 6 x ^ { 2 }$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(-4x-4+3x^{2}\right)$$

Consider -4x-4+3x^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$3x^{2}-4x-4$$

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-4 ab=3\left(-4\right)=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

Rewrite 3x^{2}-4x-4 as \left(3x^{2}-6x\right)+\left(2x-4\right).

$$\left(3x^{2}-6x\right)+\left(2x-4\right)$$

Factor out 3x in the first and 2 in the second group.

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

Factor out common term x-2 by using distributive property.

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

Rewrite the complete factored expression.

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

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$6x^{2}-8x-8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 6\left(-8\right)}}{2\times 6}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 6\left(-8\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-8\right)±\sqrt{64-24\left(-8\right)}}{2\times 6}$$

Multiply -24 times -8.

$$x=\frac{-\left(-8\right)±\sqrt{64+192}}{2\times 6}$$

Add 64 to 192.

$$x=\frac{-\left(-8\right)±\sqrt{256}}{2\times 6}$$

Take the square root of 256.

$$x=\frac{-\left(-8\right)±16}{2\times 6}$$

The opposite of -8 is 8.

$$x=\frac{8±16}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{8±16}{12}$$

Now solve the equation x=\frac{8±16}{12} when ± is plus. Add 8 to 16.

$$x=\frac{24}{12}$$

Divide 24 by 12.

$$x=2$$

Now solve the equation x=\frac{8±16}{12} when ± is minus. Subtract 16 from 8.

$$x=-\frac{8}{12}$$

Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.

$$x=-\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -\frac{2}{3} for x_{2}.

$$6x^{2}-8x-8=6\left(x-2\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$6x^{2}-8x-8=6\left(x-2\right)\left(x+\frac{2}{3}\right)$$

Add \frac{2}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-8x-8=6\left(x-2\right)\times \frac{3x+2}{3}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6x^{2}-8x-8=2\left(x-2\right)\left(3x+2\right)$$

Evaluate

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