Solve 6x^2-x-1-0

$$6 x ^ { 2 } - x - 1 - 0$$

Evaluate

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

Solution Steps

Multiply -1 and 0 to get 0.

$$6x^{2}-x-1+0$$

Add -1 and 0 to get -1.

$$6x^{2}-x-1$$

Factor

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

Solution Steps

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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-3 b=2$$

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

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

Factor out 3x in 6x^{2}-3x.

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

Factor out common term 2x-1 by using distributive property.

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

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.

$$6x^{2}-x-1=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 6\left(-1\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-1\right)±\sqrt{1-24\left(-1\right)}}{2\times 6}$$

Multiply -24 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\times 6}$$

Add 1 to 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\times 6}$$

Take the square root of 25.

$$x=\frac{-\left(-1\right)±5}{2\times 6}$$

The opposite of -1 is 1.

$$x=\frac{1±5}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{1±5}{12}$$

Now solve the equation x=\frac{1±5}{12} when ± is plus. Add 1 to 5.

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

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

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

Now solve the equation x=\frac{1±5}{12} when ± is minus. Subtract 5 from 1.

$$x=-\frac{4}{12}$$

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

$$x=-\frac{1}{3}$$

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

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

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

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

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

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

Add \frac{1}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-x-1=6\times \frac{2x-1}{2}\times \frac{3x+1}{3}$$

Multiply \frac{2x-1}{2} times \frac{3x+1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

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

Multiply 2 times 3.

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

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

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