Solve 10000-200n-8n^2

$$10000-200n-8 { n }^{ 2 }$$

Factor

$$8\left(25-n\right)\left(n+50\right)$$

Solution Steps

Factor out 8.

$$8\left(1250-25n-n^{2}\right)$$

Consider 1250-25n-n^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-n^{2}-25n+1250$$

Factor the expression by grouping. First, the expression needs to be rewritten as -n^{2}+an+bn+1250. To find a and b, set up a system to be solved.

$$a+b=-25 ab=-1250=-1250$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -1250.

$$1,-1250 2,-625 5,-250 10,-125 25,-50$$

Calculate the sum for each pair.

$$1-1250=-1249 2-625=-623 5-250=-245 10-125=-115 25-50=-25$$

The solution is the pair that gives sum -25.

$$a=25 b=-50$$

Rewrite -n^{2}-25n+1250 as \left(-n^{2}+25n\right)+\left(-50n+1250\right).

$$\left(-n^{2}+25n\right)+\left(-50n+1250\right)$$

Factor out n in the first and 50 in the second group.

$$n\left(-n+25\right)+50\left(-n+25\right)$$

Factor out common term -n+25 by using distributive property.

$$\left(-n+25\right)\left(n+50\right)$$

Rewrite the complete factored expression.

$$8\left(-n+25\right)\left(n+50\right)$$

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.

$$-8n^{2}-200n+10000=0$$

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.

$$n=\frac{-\left(-200\right)±\sqrt{\left(-200\right)^{2}-4\left(-8\right)\times 10000}}{2\left(-8\right)}$$

Square -200.

$$n=\frac{-\left(-200\right)±\sqrt{40000-4\left(-8\right)\times 10000}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$n=\frac{-\left(-200\right)±\sqrt{40000+32\times 10000}}{2\left(-8\right)}$$

Multiply 32 times 10000.

$$n=\frac{-\left(-200\right)±\sqrt{40000+320000}}{2\left(-8\right)}$$

Add 40000 to 320000.

$$n=\frac{-\left(-200\right)±\sqrt{360000}}{2\left(-8\right)}$$

Take the square root of 360000.

$$n=\frac{-\left(-200\right)±600}{2\left(-8\right)}$$

The opposite of -200 is 200.

$$n=\frac{200±600}{2\left(-8\right)}$$

Multiply 2 times -8.

$$n=\frac{200±600}{-16}$$

Now solve the equation n=\frac{200±600}{-16} when ± is plus. Add 200 to 600.

$$n=\frac{800}{-16}$$

Divide 800 by -16.

$$n=-50$$

Now solve the equation n=\frac{200±600}{-16} when ± is minus. Subtract 600 from 200.

$$n=-\frac{400}{-16}$$

Divide -400 by -16.

$$n=25$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -50 for x_{1} and 25 for x_{2}.

$$-8n^{2}-200n+10000=-8\left(n-\left(-50\right)\right)\left(n-25\right)$$

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

$$-8n^{2}-200n+10000=-8\left(n+50\right)\left(n-25\right)$$

Evaluate

$$8\left(25-n\right)\left(n+50\right)$$