Solve -16x^2+64x+192

$$-16 { x }^{ 2 } +64x+192$$

Factor

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

Solution Steps

Factor out 16.

$$16\left(-x^{2}+4x+12\right)$$

Consider -x^{2}+4x+12. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=4 ab=-12=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=6 b=-2$$

Rewrite -x^{2}+4x+12 as \left(-x^{2}+6x\right)+\left(-2x+12\right).

$$\left(-x^{2}+6x\right)+\left(-2x+12\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(x-6\right)-2\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x-2\right)$$

Rewrite the complete factored expression.

$$16\left(x-6\right)\left(-x-2\right)$$

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.

$$-16x^{2}+64x+192=0$$

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.

$$x=\frac{-64±\sqrt{64^{2}-4\left(-16\right)\times 192}}{2\left(-16\right)}$$

Square 64.

$$x=\frac{-64±\sqrt{4096-4\left(-16\right)\times 192}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-64±\sqrt{4096+64\times 192}}{2\left(-16\right)}$$

Multiply 64 times 192.

$$x=\frac{-64±\sqrt{4096+12288}}{2\left(-16\right)}$$

Add 4096 to 12288.

$$x=\frac{-64±\sqrt{16384}}{2\left(-16\right)}$$

Take the square root of 16384.

$$x=\frac{-64±128}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-64±128}{-32}$$

Now solve the equation x=\frac{-64±128}{-32} when ± is plus. Add -64 to 128.

$$x=\frac{64}{-32}$$

Divide 64 by -32.

$$x=-2$$

Now solve the equation x=\frac{-64±128}{-32} when ± is minus. Subtract 128 from -64.

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

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 -2 for x_{1} and 6 for x_{2}.

$$-16x^{2}+64x+192=-16\left(x-\left(-2\right)\right)\left(x-6\right)$$

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

$$-16x^{2}+64x+192=-16\left(x+2\right)\left(x-6\right)$$

Evaluate

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