Solve 2x^2+(-2x)+(-x)-1

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

Factor

$$2\left(x-\frac{3-\sqrt{17}}{4}\right)\left(x-\frac{\sqrt{17}+3}{4}\right)$$

Steps Using the Quadratic Formula

Combine -2x and -x to get -3x.

$$factor(2x^{2}-3x-1)$$

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.

$$2x^{2}-3x-1=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-1\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+8}}{2\times 2}$$

Add 9 to 8.

$$x=\frac{-\left(-3\right)±\sqrt{17}}{2\times 2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±\sqrt{17}}{4}$$

Now solve the equation x=\frac{3±\sqrt{17}}{4} when ± is plus. Add 3 to \sqrt{17}.

$$x=\frac{\sqrt{17}+3}{4}$$

Now solve the equation x=\frac{3±\sqrt{17}}{4} when ± is minus. Subtract \sqrt{17} from 3.

$$x=\frac{3-\sqrt{17}}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3+\sqrt{17}}{4} for x_{1} and \frac{3-\sqrt{17}}{4} for x_{2}.

$$2x^{2}-3x-1=2\left(x-\frac{\sqrt{17}+3}{4}\right)\left(x-\frac{3-\sqrt{17}}{4}\right)$$

Evaluate

$$2x^{2}-3x-1$$

Solution Steps

Combine -2x and -x to get -3x.

$$2x^{2}-3x-1$$