Solve f(x)=-16x^2+96x

$$f ( x ) = - 16 x ^ { 2 } + 96 x$$

Factor

$$16x\left(6-x\right)$$

Solution Steps

Factor out 16.

$$16\left(-x^{2}+6x\right)$$

Consider -x^{2}+6x. Factor out x.

$$x\left(-x+6\right)$$

Rewrite the complete factored expression.

$$16x\left(-x+6\right)$$

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.

$$-16x^{2}+96x=0$$

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.

$$x=\frac{-96±\sqrt{96^{2}}}{2\left(-16\right)}$$

Take the square root of 96^{2}.

$$x=\frac{-96±96}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-96±96}{-32}$$

Now solve the equation x=\frac{-96±96}{-32} when ± is plus. Add -96 to 96.

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

Divide 0 by -32.

$$x=0$$

Now solve the equation x=\frac{-96±96}{-32} when ± is minus. Subtract 96 from -96.

$$x=-\frac{192}{-32}$$

Divide -192 by -32.

$$x=6$$

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 6 for x_{2}.

$$-16x^{2}+96x=-16x\left(x-6\right)$$

Evaluate

$$16x\left(6-x\right)$$