Solve x^2+105x-1150

$${ x }^{ 2 } +105x-1150$$

Factor

$$\left(x-10\right)\left(x+115\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-1150. To find a and b, set up a system to be solved.

$$a+b=105 ab=1\left(-1150\right)=-1150$$

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 -1150.

$$-1,1150 -2,575 -5,230 -10,115 -23,50 -25,46$$

Calculate the sum for each pair.

$$-1+1150=1149 -2+575=573 -5+230=225 -10+115=105 -23+50=27 -25+46=21$$

The solution is the pair that gives sum 105.

$$a=-10 b=115$$

Rewrite x^{2}+105x-1150 as \left(x^{2}-10x\right)+\left(115x-1150\right).

$$\left(x^{2}-10x\right)+\left(115x-1150\right)$$

Factor out x in the first and 115 in the second group.

$$x\left(x-10\right)+115\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x+115\right)$$

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.

$$x^{2}+105x-1150=0$$

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.

$$x=\frac{-105±\sqrt{105^{2}-4\left(-1150\right)}}{2}$$

Square 105.

$$x=\frac{-105±\sqrt{11025-4\left(-1150\right)}}{2}$$

Multiply -4 times -1150.

$$x=\frac{-105±\sqrt{11025+4600}}{2}$$

Add 11025 to 4600.

$$x=\frac{-105±\sqrt{15625}}{2}$$

Take the square root of 15625.

$$x=\frac{-105±125}{2}$$

Now solve the equation x=\frac{-105±125}{2} when ± is plus. Add -105 to 125.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{-105±125}{2} when ± is minus. Subtract 125 from -105.

$$x=-\frac{230}{2}$$

Divide -230 by 2.

$$x=-115$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10 for x_{1} and -115 for x_{2}.

$$x^{2}+105x-1150=\left(x-10\right)\left(x-\left(-115\right)\right)$$

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

$$x^{2}+105x-1150=\left(x-10\right)\left(x+115\right)$$

Evaluate

$$\left(x-10\right)\left(x+115\right)$$