Solve 16+6x-x^2

$$16 + 6 x - x ^ { 2 }$$

Factor

$$-\left(x-8\right)\left(x+2\right)$$

Solution Steps

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

$$-x^{2}+6x+16$$

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=6 ab=-16=-16$$

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 -16.

$$-1,16 -2,8 -4,4$$

Calculate the sum for each pair.

$$-1+16=15 -2+8=6 -4+4=0$$

The solution is the pair that gives sum 6.

$$a=8 b=-2$$

Rewrite -x^{2}+6x+16 as \left(-x^{2}+8x\right)+\left(-2x+16\right).

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

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

$$-x\left(x-8\right)-2\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(-x-2\right)$$

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.

$$-x^{2}+6x+16=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\times 16}}{2\left(-1\right)}$$

Multiply 4 times 16.

$$x=\frac{-6±\sqrt{36+64}}{2\left(-1\right)}$$

Add 36 to 64.

$$x=\frac{-6±\sqrt{100}}{2\left(-1\right)}$$

Take the square root of 100.

$$x=\frac{-6±10}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±10}{-2}$$

Now solve the equation x=\frac{-6±10}{-2} when ± is plus. Add -6 to 10.

$$x=\frac{4}{-2}$$

Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{-6±10}{-2} when ± is minus. Subtract 10 from -6.

$$x=-\frac{16}{-2}$$

Divide -16 by -2.

$$x=8$$

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 8 for x_{2}.

$$-x^{2}+6x+16=-\left(x-\left(-2\right)\right)\left(x-8\right)$$

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

$$-x^{2}+6x+16=-\left(x+2\right)\left(x-8\right)$$

Evaluate

$$-\left(x-8\right)\left(x+2\right)$$