Solve 18r^2+132r+240

$$18 r ^ { 2 } + 132 r + 240$$

Factor

$$6\left(r+4\right)\left(3r+10\right)$$

Solution Steps

Factor out 6.

$$6\left(3r^{2}+22r+40\right)$$

Consider 3r^{2}+22r+40. Factor the expression by grouping. First, the expression needs to be rewritten as 3r^{2}+ar+br+40. To find a and b, set up a system to be solved.

$$a+b=22 ab=3\times 40=120$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 22.

$$a=10 b=12$$

Rewrite 3r^{2}+22r+40 as \left(3r^{2}+10r\right)+\left(12r+40\right).

$$\left(3r^{2}+10r\right)+\left(12r+40\right)$$

Factor out r in the first and 4 in the second group.

$$r\left(3r+10\right)+4\left(3r+10\right)$$

Factor out common term 3r+10 by using distributive property.

$$\left(3r+10\right)\left(r+4\right)$$

Rewrite the complete factored expression.

$$6\left(3r+10\right)\left(r+4\right)$$

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.

$$18r^{2}+132r+240=0$$

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.

$$r=\frac{-132±\sqrt{132^{2}-4\times 18\times 240}}{2\times 18}$$

Square 132.

$$r=\frac{-132±\sqrt{17424-4\times 18\times 240}}{2\times 18}$$

Multiply -4 times 18.

$$r=\frac{-132±\sqrt{17424-72\times 240}}{2\times 18}$$

Multiply -72 times 240.

$$r=\frac{-132±\sqrt{17424-17280}}{2\times 18}$$

Add 17424 to -17280.

$$r=\frac{-132±\sqrt{144}}{2\times 18}$$

Take the square root of 144.

$$r=\frac{-132±12}{2\times 18}$$

Multiply 2 times 18.

$$r=\frac{-132±12}{36}$$

Now solve the equation r=\frac{-132±12}{36} when ± is plus. Add -132 to 12.

$$r=-\frac{120}{36}$$

Reduce the fraction \frac{-120}{36} to lowest terms by extracting and canceling out 12.

$$r=-\frac{10}{3}$$

Now solve the equation r=\frac{-132±12}{36} when ± is minus. Subtract 12 from -132.

$$r=-\frac{144}{36}$$

Divide -144 by 36.

$$r=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{10}{3} for x_{1} and -4 for x_{2}.

$$18r^{2}+132r+240=18\left(r-\left(-\frac{10}{3}\right)\right)\left(r-\left(-4\right)\right)$$

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

$$18r^{2}+132r+240=18\left(r+\frac{10}{3}\right)\left(r+4\right)$$

Add \frac{10}{3} to r by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18r^{2}+132r+240=18\times \frac{3r+10}{3}\left(r+4\right)$$

Cancel out 3, the greatest common factor in 18 and 3.

$$18r^{2}+132r+240=6\left(3r+10\right)\left(r+4\right)$$

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 18

$$x ^ 2 +\frac{22}{3}x +\frac{40}{3} = 0$$

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

$$r + s = -\frac{22}{3} rs = \frac{40}{3}$$

Two numbers r and s sum up to -\frac{22}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{22}{3} = -\frac{11}{3}. 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>

$$r = -\frac{11}{3} - u s = -\frac{11}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{40}{3}

$$(-\frac{11}{3} - u) (-\frac{11}{3} + u) = \frac{40}{3}$$

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

$$\frac{121}{9} - u^2 = \frac{40}{3}$$

Simplify the expression by subtracting \frac{121}{9} on both sides

$$-u^2 = \frac{40}{3}-\frac{121}{9} = -\frac{1}{9}$$

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

$$u^2 = \frac{1}{9} u = \pm\sqrt{\frac{1}{9}} = \pm \frac{1}{3}$$

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

$$r =-\frac{11}{3} - \frac{1}{3} = -4.000 s = -\frac{11}{3} + \frac{1}{3} = -3.333$$

Evaluate

$$6\left(r+4\right)\left(3r+10\right)$$