Solve -2800+160x-x^2

$$-2800+160x- { x }^{ 2 }$$

Factor

$$\left(20-x\right)\left(x-140\right)$$

Solution Steps

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

$$-x^{2}+160x-2800$$

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-2800. To find a and b, set up a system to be solved.

$$a+b=160 ab=-\left(-2800\right)=2800$$

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 2800.

$$1,2800 2,1400 4,700 5,560 7,400 8,350 10,280 14,200 16,175 20,140 25,112 28,100 35,80 40,70 50,56$$

Calculate the sum for each pair.

$$1+2800=2801 2+1400=1402 4+700=704 5+560=565 7+400=407 8+350=358 10+280=290 14+200=214 16+175=191 20+140=160 25+112=137 28+100=128 35+80=115 40+70=110 50+56=106$$

The solution is the pair that gives sum 160.

$$a=140 b=20$$

Rewrite -x^{2}+160x-2800 as \left(-x^{2}+140x\right)+\left(20x-2800\right).

$$\left(-x^{2}+140x\right)+\left(20x-2800\right)$$

Factor out -x in the first and 20 in the second group.

$$-x\left(x-140\right)+20\left(x-140\right)$$

Factor out common term x-140 by using distributive property.

$$\left(x-140\right)\left(-x+20\right)$$

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.

$$-x^{2}+160x-2800=0$$

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.

$$x=\frac{-160±\sqrt{160^{2}-4\left(-1\right)\left(-2800\right)}}{2\left(-1\right)}$$

Square 160.

$$x=\frac{-160±\sqrt{25600-4\left(-1\right)\left(-2800\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-160±\sqrt{25600+4\left(-2800\right)}}{2\left(-1\right)}$$

Multiply 4 times -2800.

$$x=\frac{-160±\sqrt{25600-11200}}{2\left(-1\right)}$$

Add 25600 to -11200.

$$x=\frac{-160±\sqrt{14400}}{2\left(-1\right)}$$

Take the square root of 14400.

$$x=\frac{-160±120}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-160±120}{-2}$$

Now solve the equation x=\frac{-160±120}{-2} when ± is plus. Add -160 to 120.

$$x=-\frac{40}{-2}$$

Divide -40 by -2.

$$x=20$$

Now solve the equation x=\frac{-160±120}{-2} when ± is minus. Subtract 120 from -160.

$$x=-\frac{280}{-2}$$

Divide -280 by -2.

$$x=140$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 20 for x_{1} and 140 for x_{2}.

$$-x^{2}+160x-2800=-\left(x-20\right)\left(x-140\right)$$

Evaluate

$$\left(20-x\right)\left(x-140\right)$$