Solve -16x^2+128x

$$-16 { x }^{ 2 } +128x$$

Factor

$$16x\left(8-x\right)$$

Solution Steps

Factor out 16.

$$16\left(-x^{2}+8x\right)$$

Consider -x^{2}+8x. Factor out x.

$$x\left(-x+8\right)$$

Rewrite the complete factored expression.

$$16x\left(-x+8\right)$$

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.

$$-16x^{2}+128x=0$$

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.

$$x=\frac{-128±\sqrt{128^{2}}}{2\left(-16\right)}$$

Take the square root of 128^{2}.

$$x=\frac{-128±128}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-128±128}{-32}$$

Now solve the equation x=\frac{-128±128}{-32} when ± is plus. Add -128 to 128.

$$x=\frac{0}{-32}$$

Divide 0 by -32.

$$x=0$$

Now solve the equation x=\frac{-128±128}{-32} when ± is minus. Subtract 128 from -128.

$$x=-\frac{256}{-32}$$

Divide -256 by -32.

$$x=8$$

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 8 for x_{2}.

$$-16x^{2}+128x=-16x\left(x-8\right)$$

Evaluate

$$16x\left(8-x\right)$$