Solve x^2-50x-5000

$${ x }^{ 2 } -50x-5000$$

Factor

$$\left(x-100\right)\left(x+50\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-5000. To find a and b, set up a system to be solved.

$$a+b=-50 ab=1\left(-5000\right)=-5000$$

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 -5000.

$$1,-5000 2,-2500 4,-1250 5,-1000 8,-625 10,-500 20,-250 25,-200 40,-125 50,-100$$

Calculate the sum for each pair.

$$1-5000=-4999 2-2500=-2498 4-1250=-1246 5-1000=-995 8-625=-617 10-500=-490 20-250=-230 25-200=-175 40-125=-85 50-100=-50$$

The solution is the pair that gives sum -50.

$$a=-100 b=50$$

Rewrite x^{2}-50x-5000 as \left(x^{2}-100x\right)+\left(50x-5000\right).

$$\left(x^{2}-100x\right)+\left(50x-5000\right)$$

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

$$x\left(x-100\right)+50\left(x-100\right)$$

Factor out common term x-100 by using distributive property.

$$\left(x-100\right)\left(x+50\right)$$

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.

$$x^{2}-50x-5000=0$$

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.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\left(-5000\right)}}{2}$$

Square -50.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\left(-5000\right)}}{2}$$

Multiply -4 times -5000.

$$x=\frac{-\left(-50\right)±\sqrt{2500+20000}}{2}$$

Add 2500 to 20000.

$$x=\frac{-\left(-50\right)±\sqrt{22500}}{2}$$

Take the square root of 22500.

$$x=\frac{-\left(-50\right)±150}{2}$$

The opposite of -50 is 50.

$$x=\frac{50±150}{2}$$

Now solve the equation x=\frac{50±150}{2} when ± is plus. Add 50 to 150.

$$x=\frac{200}{2}$$

Divide 200 by 2.

$$x=100$$

Now solve the equation x=\frac{50±150}{2} when ± is minus. Subtract 150 from 50.

$$x=-\frac{100}{2}$$

Divide -100 by 2.

$$x=-50$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 100 for x_{1} and -50 for x_{2}.

$$x^{2}-50x-5000=\left(x-100\right)\left(x-\left(-50\right)\right)$$

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

$$x^{2}-50x-5000=\left(x-100\right)\left(x+50\right)$$

Evaluate

$$\left(x-100\right)\left(x+50\right)$$