Solve 6x^2-3x-3

$$6 { x }^{ 2 } -3x-3$$

Factor

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

Solution Steps

Factor out 3.

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

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

$$a+b=-1 ab=2\left(-1\right)=-2$$

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. The only such pair is the system solution.

$$a=-2 b=1$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$6x^{2}-3x-3=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 6\left(-3\right)}}{2\times 6}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 6\left(-3\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-3\right)±\sqrt{9-24\left(-3\right)}}{2\times 6}$$

Multiply -24 times -3.

$$x=\frac{-\left(-3\right)±\sqrt{9+72}}{2\times 6}$$

Add 9 to 72.

$$x=\frac{-\left(-3\right)±\sqrt{81}}{2\times 6}$$

Take the square root of 81.

$$x=\frac{-\left(-3\right)±9}{2\times 6}$$

The opposite of -3 is 3.

$$x=\frac{3±9}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{3±9}{12}$$

Now solve the equation x=\frac{3±9}{12} when ± is plus. Add 3 to 9.

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

Divide 12 by 12.

$$x=1$$

Now solve the equation x=\frac{3±9}{12} when ± is minus. Subtract 9 from 3.

$$x=-\frac{6}{12}$$

Reduce the fraction \frac{-6}{12} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{2}$$

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

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

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

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

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 2, the greatest common factor in 6 and 2.

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

Evaluate

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