Solve 1000x-2x^2

$$1000x-2 { x }^{ 2 }$$

Factor

$$2x\left(500-x\right)$$

Solution Steps

Factor out 2.

$$2\left(500x-x^{2}\right)$$

Consider 500x-x^{2}. Factor out x.

$$x\left(500-x\right)$$

Rewrite the complete factored expression.

$$2x\left(-x+500\right)$$

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.

$$-2x^{2}+1000x=0$$

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.

$$x=\frac{-1000±\sqrt{1000^{2}}}{2\left(-2\right)}$$

Take the square root of 1000^{2}.

$$x=\frac{-1000±1000}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-1000±1000}{-4}$$

Now solve the equation x=\frac{-1000±1000}{-4} when ± is plus. Add -1000 to 1000.

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

Divide 0 by -4.

$$x=0$$

Now solve the equation x=\frac{-1000±1000}{-4} when ± is minus. Subtract 1000 from -1000.

$$x=-\frac{2000}{-4}$$

Divide -2000 by -4.

$$x=500$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and 500 for x_{2}.

$$-2x^{2}+1000x=-2x\left(x-500\right)$$

Evaluate

$$2x\left(500-x\right)$$