Solve 10x-x^2-16

$$10 x - x ^ { 2 } - 16$$

Factor

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

Solution Steps

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

$$-x^{2}+10x-16$$

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.

$$a+b=10 ab=-\left(-16\right)=16$$

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 16.

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

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 10.

$$a=8 b=2$$

Rewrite -x^{2}+10x-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.

$$-x^{2}+10x-16=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-16\right)}}{2\left(-1\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-16\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-10±\sqrt{100+4\left(-16\right)}}{2\left(-1\right)}$$

Multiply 4 times -16.

$$x=\frac{-10±\sqrt{100-64}}{2\left(-1\right)}$$

Add 100 to -64.

$$x=\frac{-10±\sqrt{36}}{2\left(-1\right)}$$

Take the square root of 36.

$$x=\frac{-10±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-10±6}{-2}$$

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

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

Divide -4 by -2.

$$x=2$$

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

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

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}.

$$-x^{2}+10x-16=-\left(x-2\right)\left(x-8\right)$$

Evaluate

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