Solve x^2+90x-5200

$${ x }^{ 2 } +90x-5200$$

Factor

$$\left(x-40\right)\left(x+130\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-5200. To find a and b, set up a system to be solved.

$$a+b=90 ab=1\left(-5200\right)=-5200$$

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 -5200.

$$-1,5200 -2,2600 -4,1300 -5,1040 -8,650 -10,520 -13,400 -16,325 -20,260 -25,208 -26,200 -40,130 -50,104 -52,100 -65,80$$

Calculate the sum for each pair.

$$-1+5200=5199 -2+2600=2598 -4+1300=1296 -5+1040=1035 -8+650=642 -10+520=510 -13+400=387 -16+325=309 -20+260=240 -25+208=183 -26+200=174 -40+130=90 -50+104=54 -52+100=48 -65+80=15$$

The solution is the pair that gives sum 90.

$$a=-40 b=130$$

Rewrite x^{2}+90x-5200 as \left(x^{2}-40x\right)+\left(130x-5200\right).

$$\left(x^{2}-40x\right)+\left(130x-5200\right)$$

Factor out x in the first and 130 in the second group.

$$x\left(x-40\right)+130\left(x-40\right)$$

Factor out common term x-40 by using distributive property.

$$\left(x-40\right)\left(x+130\right)$$

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.

$$x^{2}+90x-5200=0$$

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.

$$x=\frac{-90±\sqrt{90^{2}-4\left(-5200\right)}}{2}$$

Square 90.

$$x=\frac{-90±\sqrt{8100-4\left(-5200\right)}}{2}$$

Multiply -4 times -5200.

$$x=\frac{-90±\sqrt{8100+20800}}{2}$$

Add 8100 to 20800.

$$x=\frac{-90±\sqrt{28900}}{2}$$

Take the square root of 28900.

$$x=\frac{-90±170}{2}$$

Now solve the equation x=\frac{-90±170}{2} when ± is plus. Add -90 to 170.

$$x=\frac{80}{2}$$

Divide 80 by 2.

$$x=40$$

Now solve the equation x=\frac{-90±170}{2} when ± is minus. Subtract 170 from -90.

$$x=-\frac{260}{2}$$

Divide -260 by 2.

$$x=-130$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 40 for x_{1} and -130 for x_{2}.

$$x^{2}+90x-5200=\left(x-40\right)\left(x-\left(-130\right)\right)$$

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

$$x^{2}+90x-5200=\left(x-40\right)\left(x+130\right)$$

Evaluate

$$\left(x-40\right)\left(x+130\right)$$