Solve h(x)=-5x^2+80x

$$h ( x ) = - 5 x ^ { 2 } + 80 x$$

Factor

$$5x\left(16-x\right)$$

Solution Steps

Factor out 5.

$$5\left(-x^{2}+16x\right)$$

Consider -x^{2}+16x. Factor out x.

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

Rewrite the complete factored expression.

$$5x\left(-x+16\right)$$

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.

$$-5x^{2}+80x=0$$

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.

$$x=\frac{-80±\sqrt{80^{2}}}{2\left(-5\right)}$$

Take the square root of 80^{2}.

$$x=\frac{-80±80}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-80±80}{-10}$$

Now solve the equation x=\frac{-80±80}{-10} when ± is plus. Add -80 to 80.

$$x=\frac{0}{-10}$$

Divide 0 by -10.

$$x=0$$

Now solve the equation x=\frac{-80±80}{-10} when ± is minus. Subtract 80 from -80.

$$x=-\frac{160}{-10}$$

Divide -160 by -10.

$$x=16$$

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 16 for x_{2}.

$$-5x^{2}+80x=-5x\left(x-16\right)$$

Evaluate

$$5x\left(16-x\right)$$