Solve 18r^2+9r-20

$$18 r ^ { 2 } + 9 r - 20$$

Factor

$$\left(6r-5\right)\left(3r+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 18r^{2}+ar+br-20. To find a and b, set up a system to be solved.

$$a+b=9 ab=18\left(-20\right)=-360$$

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 -360.

$$-1,360 -2,180 -3,120 -4,90 -5,72 -6,60 -8,45 -9,40 -10,36 -12,30 -15,24 -18,20$$

Calculate the sum for each pair.

$$-1+360=359 -2+180=178 -3+120=117 -4+90=86 -5+72=67 -6+60=54 -8+45=37 -9+40=31 -10+36=26 -12+30=18 -15+24=9 -18+20=2$$

The solution is the pair that gives sum 9.

$$a=-15 b=24$$

Rewrite 18r^{2}+9r-20 as \left(18r^{2}-15r\right)+\left(24r-20\right).

$$\left(18r^{2}-15r\right)+\left(24r-20\right)$$

Factor out 3r in the first and 4 in the second group.

$$3r\left(6r-5\right)+4\left(6r-5\right)$$

Factor out common term 6r-5 by using distributive property.

$$\left(6r-5\right)\left(3r+4\right)$$

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.

$$18r^{2}+9r-20=0$$

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.

$$r=\frac{-9±\sqrt{9^{2}-4\times 18\left(-20\right)}}{2\times 18}$$

Square 9.

$$r=\frac{-9±\sqrt{81-4\times 18\left(-20\right)}}{2\times 18}$$

Multiply -4 times 18.

$$r=\frac{-9±\sqrt{81-72\left(-20\right)}}{2\times 18}$$

Multiply -72 times -20.

$$r=\frac{-9±\sqrt{81+1440}}{2\times 18}$$

Add 81 to 1440.

$$r=\frac{-9±\sqrt{1521}}{2\times 18}$$

Take the square root of 1521.

$$r=\frac{-9±39}{2\times 18}$$

Multiply 2 times 18.

$$r=\frac{-9±39}{36}$$

Now solve the equation r=\frac{-9±39}{36} when ± is plus. Add -9 to 39.

$$r=\frac{30}{36}$$

Reduce the fraction \frac{30}{36} to lowest terms by extracting and canceling out 6.

$$r=\frac{5}{6}$$

Now solve the equation r=\frac{-9±39}{36} when ± is minus. Subtract 39 from -9.

$$r=-\frac{48}{36}$$

Reduce the fraction \frac{-48}{36} to lowest terms by extracting and canceling out 12.

$$r=-\frac{4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{6} for x_{1} and -\frac{4}{3} for x_{2}.

$$18r^{2}+9r-20=18\left(r-\frac{5}{6}\right)\left(r-\left(-\frac{4}{3}\right)\right)$$

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

$$18r^{2}+9r-20=18\left(r-\frac{5}{6}\right)\left(r+\frac{4}{3}\right)$$

Subtract \frac{5}{6} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18r^{2}+9r-20=18\times \frac{6r-5}{6}\left(r+\frac{4}{3}\right)$$

Add \frac{4}{3} to r by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18r^{2}+9r-20=18\times \frac{6r-5}{6}\times \frac{3r+4}{3}$$

Multiply \frac{6r-5}{6} times \frac{3r+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18r^{2}+9r-20=18\times \frac{\left(6r-5\right)\left(3r+4\right)}{6\times 3}$$

Multiply 6 times 3.

$$18r^{2}+9r-20=18\times \frac{\left(6r-5\right)\left(3r+4\right)}{18}$$

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

$$18r^{2}+9r-20=\left(6r-5\right)\left(3r+4\right)$$

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

$$x ^ 2 +\frac{1}{2}x -\frac{10}{9} = 0$$

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

$$r + s = -\frac{1}{2} rs = -\frac{10}{9}$$

Two numbers r and s sum up to -\frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{2} = -\frac{1}{4}. 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>

$$r = -\frac{1}{4} - u s = -\frac{1}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{10}{9}

$$(-\frac{1}{4} - u) (-\frac{1}{4} + u) = -\frac{10}{9}$$

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

$$\frac{1}{16} - u^2 = -\frac{10}{9}$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -\frac{10}{9}-\frac{1}{16} = -\frac{169}{144}$$

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

$$u^2 = \frac{169}{144} u = \pm\sqrt{\frac{169}{144}} = \pm \frac{13}{12}$$

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

$$r =-\frac{1}{4} - \frac{13}{12} = -1.333 s = -\frac{1}{4} + \frac{13}{12} = 0.833$$

Evaluate

$$\left(6r-5\right)\left(3r+4\right)$$