Solve x^2+201x-5400

$${ x }^{ 2 } +201x-5400$$

Factor

$$\left(x-24\right)\left(x+225\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-5400. To find a and b, set up a system to be solved.

$$a+b=201 ab=1\left(-5400\right)=-5400$$

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 -5400.

$$-1,5400 -2,2700 -3,1800 -4,1350 -5,1080 -6,900 -8,675 -9,600 -10,540 -12,450 -15,360 -18,300 -20,270 -24,225 -25,216 -27,200 -30,180 -36,150 -40,135 -45,120 -50,108 -54,100 -60,90 -72,75$$

Calculate the sum for each pair.

$$-1+5400=5399 -2+2700=2698 -3+1800=1797 -4+1350=1346 -5+1080=1075 -6+900=894 -8+675=667 -9+600=591 -10+540=530 -12+450=438 -15+360=345 -18+300=282 -20+270=250 -24+225=201 -25+216=191 -27+200=173 -30+180=150 -36+150=114 -40+135=95 -45+120=75 -50+108=58 -54+100=46 -60+90=30 -72+75=3$$

The solution is the pair that gives sum 201.

$$a=-24 b=225$$

Rewrite x^{2}+201x-5400 as \left(x^{2}-24x\right)+\left(225x-5400\right).

$$\left(x^{2}-24x\right)+\left(225x-5400\right)$$

Factor out x in the first and 225 in the second group.

$$x\left(x-24\right)+225\left(x-24\right)$$

Factor out common term x-24 by using distributive property.

$$\left(x-24\right)\left(x+225\right)$$

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.

$$x^{2}+201x-5400=0$$

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.

$$x=\frac{-201±\sqrt{201^{2}-4\left(-5400\right)}}{2}$$

Square 201.

$$x=\frac{-201±\sqrt{40401-4\left(-5400\right)}}{2}$$

Multiply -4 times -5400.

$$x=\frac{-201±\sqrt{40401+21600}}{2}$$

Add 40401 to 21600.

$$x=\frac{-201±\sqrt{62001}}{2}$$

Take the square root of 62001.

$$x=\frac{-201±249}{2}$$

Now solve the equation x=\frac{-201±249}{2} when ± is plus. Add -201 to 249.

$$x=\frac{48}{2}$$

Divide 48 by 2.

$$x=24$$

Now solve the equation x=\frac{-201±249}{2} when ± is minus. Subtract 249 from -201.

$$x=-\frac{450}{2}$$

Divide -450 by 2.

$$x=-225$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 24 for x_{1} and -225 for x_{2}.

$$x^{2}+201x-5400=\left(x-24\right)\left(x-\left(-225\right)\right)$$

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

$$x^{2}+201x-5400=\left(x-24\right)\left(x+225\right)$$

Evaluate

$$\left(x-24\right)\left(x+225\right)$$