Solve -2x^2-x

$$- 2 x ^ { 2 } - x$$

Factor

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

Solution Steps

Factor out x.

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

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.

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

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.

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

Take the square root of 1.

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

The opposite of -1 is 1.

$$x=\frac{1±1}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{1±1}{-4}$$

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

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

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±1}{-4} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

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 0 for x_{2}.

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

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

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

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

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

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

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

Evaluate

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