Solve -x^2-4x

$$- x ^ { 2 } - 4 x$$

Factor

$$-x\left(x+4\right)$$

Solution Steps

Factor out x.

$$x\left(-x-4\right)$$

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.

$$-x^{2}-4x=0$$

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.

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

Take the square root of \left(-4\right)^{2}.

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

The opposite of -4 is 4.

$$x=\frac{4±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±4}{-2}$$

Now solve the equation x=\frac{4±4}{-2} when ± is plus. Add 4 to 4.

$$x=\frac{8}{-2}$$

Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{4±4}{-2} when ± is minus. Subtract 4 from 4.

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4 for x_{1} and 0 for x_{2}.

$$-x^{2}-4x=-\left(x-\left(-4\right)\right)x$$

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

$$-x^{2}-4x=-\left(x+4\right)x$$

Evaluate

$$-x\left(x+4\right)$$