Solve (2x^2+x-1)

$$( 2 x ^ { 2 } + x - 1 )$$

Factor

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

Solution Steps

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 positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=2$$

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

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

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

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

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

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

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.

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

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.

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

Square 1.

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

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$x=\frac{-1±\sqrt{1+8}}{2\times 2}$$

Add 1 to 8.

$$x=\frac{-1±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$x=\frac{-1±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±3}{4}$$

Now solve the equation x=\frac{-1±3}{4} when ± is plus. Add -1 to 3.

$$x=\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-1±3}{4} when ± is minus. Subtract 3 from -1.

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

Divide -4 by 4.

$$x=-1$$

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 -1 for x_{2}.

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

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

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

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

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

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

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

Evaluate

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