Solve 3n^2+603n-16200

$$3 n ^ { 2 } + 603 n - 16200$$

Factor

$$3\left(n-24\right)\left(n+225\right)$$

Solution Steps

Factor out 3.

$$3\left(n^{2}+201n-5400\right)$$

Consider n^{2}+201n-5400. Factor the expression by grouping. First, the expression needs to be rewritten as n^{2}+an+bn-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 n^{2}+201n-5400 as \left(n^{2}-24n\right)+\left(225n-5400\right).

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

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

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

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

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

Rewrite the complete factored expression.

$$3\left(n-24\right)\left(n+225\right)$$

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.

$$3n^{2}+603n-16200=0$$

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.

$$n=\frac{-603±\sqrt{603^{2}-4\times 3\left(-16200\right)}}{2\times 3}$$

Square 603.

$$n=\frac{-603±\sqrt{363609-4\times 3\left(-16200\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-603±\sqrt{363609-12\left(-16200\right)}}{2\times 3}$$

Multiply -12 times -16200.

$$n=\frac{-603±\sqrt{363609+194400}}{2\times 3}$$

Add 363609 to 194400.

$$n=\frac{-603±\sqrt{558009}}{2\times 3}$$

Take the square root of 558009.

$$n=\frac{-603±747}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{-603±747}{6}$$

Now solve the equation n=\frac{-603±747}{6} when ± is plus. Add -603 to 747.

$$n=\frac{144}{6}$$

Divide 144 by 6.

$$n=24$$

Now solve the equation n=\frac{-603±747}{6} when ± is minus. Subtract 747 from -603.

$$n=-\frac{1350}{6}$$

Divide -1350 by 6.

$$n=-225$$

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}.

$$3n^{2}+603n-16200=3\left(n-24\right)\left(n-\left(-225\right)\right)$$

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

$$3n^{2}+603n-16200=3\left(n-24\right)\left(n+225\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 3

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

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -201 rs = -5400$$

Two numbers r and s sum up to -201 exactly when the average of the two numbers is \frac{1}{2}*-201 = -\frac{201}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{201}{2} - u s = -\frac{201}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -5400

$$(-\frac{201}{2} - u) (-\frac{201}{2} + u) = -5400$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{40401}{4} - u^2 = -5400$$

Simplify the expression by subtracting \frac{40401}{4} on both sides

$$-u^2 = -5400-\frac{40401}{4} = -\frac{62001}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{62001}{4} u = \pm\sqrt{\frac{62001}{4}} = \pm \frac{249}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{201}{2} - \frac{249}{2} = -225 s = -\frac{201}{2} + \frac{249}{2} = 24$$

Evaluate

$$3\left(n-24\right)\left(n+225\right)$$