Solve 16x-x^2-48

$$16 x - x ^ { 2 } - 48$$

Factor

$$\left(4-x\right)\left(x-12\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+16x-48$$

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-48. To find a and b, set up a system to be solved.

$$a+b=16 ab=-\left(-48\right)=48$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 48.

$$1,48 2,24 3,16 4,12 6,8$$

Calculate the sum for each pair.

$$1+48=49 2+24=26 3+16=19 4+12=16 6+8=14$$

The solution is the pair that gives sum 16.

$$a=12 b=4$$

Rewrite -x^{2}+16x-48 as \left(-x^{2}+12x\right)+\left(4x-48\right).

$$\left(-x^{2}+12x\right)+\left(4x-48\right)$$

Factor out -x in the first and 4 in the second group.

$$-x\left(x-12\right)+4\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(-x+4\right)$$

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.

$$-x^{2}+16x-48=0$$

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.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-1\right)\left(-48\right)}}{2\left(-1\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-1\right)\left(-48\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-16±\sqrt{256+4\left(-48\right)}}{2\left(-1\right)}$$

Multiply 4 times -48.

$$x=\frac{-16±\sqrt{256-192}}{2\left(-1\right)}$$

Add 256 to -192.

$$x=\frac{-16±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of 64.

$$x=\frac{-16±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-16±8}{-2}$$

Now solve the equation x=\frac{-16±8}{-2} when ± is plus. Add -16 to 8.

$$x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x=4$$

Now solve the equation x=\frac{-16±8}{-2} when ± is minus. Subtract 8 from -16.

$$x=-\frac{24}{-2}$$

Divide -24 by -2.

$$x=12$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and 12 for x_{2}.

$$-x^{2}+16x-48=-\left(x-4\right)\left(x-12\right)$$

Evaluate

$$\left(4-x\right)\left(x-12\right)$$