Solve x^2+10x-1200

$${ x }^{ 2 } +10x-1200$$

Factor

$$\left(x-30\right)\left(x+40\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-1200. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\left(-1200\right)=-1200$$

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 -1200.

$$-1,1200 -2,600 -3,400 -4,300 -5,240 -6,200 -8,150 -10,120 -12,100 -15,80 -16,75 -20,60 -24,50 -25,48 -30,40$$

Calculate the sum for each pair.

$$-1+1200=1199 -2+600=598 -3+400=397 -4+300=296 -5+240=235 -6+200=194 -8+150=142 -10+120=110 -12+100=88 -15+80=65 -16+75=59 -20+60=40 -24+50=26 -25+48=23 -30+40=10$$

The solution is the pair that gives sum 10.

$$a=-30 b=40$$

Rewrite x^{2}+10x-1200 as \left(x^{2}-30x\right)+\left(40x-1200\right).

$$\left(x^{2}-30x\right)+\left(40x-1200\right)$$

Factor out x in the first and 40 in the second group.

$$x\left(x-30\right)+40\left(x-30\right)$$

Factor out common term x-30 by using distributive property.

$$\left(x-30\right)\left(x+40\right)$$

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.

$$x^{2}+10x-1200=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1200\right)}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-1200\right)}}{2}$$

Multiply -4 times -1200.

$$x=\frac{-10±\sqrt{100+4800}}{2}$$

Add 100 to 4800.

$$x=\frac{-10±\sqrt{4900}}{2}$$

Take the square root of 4900.

$$x=\frac{-10±70}{2}$$

Now solve the equation x=\frac{-10±70}{2} when ± is plus. Add -10 to 70.

$$x=\frac{60}{2}$$

Divide 60 by 2.

$$x=30$$

Now solve the equation x=\frac{-10±70}{2} when ± is minus. Subtract 70 from -10.

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

Divide -80 by 2.

$$x=-40$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 30 for x_{1} and -40 for x_{2}.

$$x^{2}+10x-1200=\left(x-30\right)\left(x-\left(-40\right)\right)$$

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

$$x^{2}+10x-1200=\left(x-30\right)\left(x+40\right)$$

Evaluate

$$\left(x-30\right)\left(x+40\right)$$