Solve 36r^2-24r-5

$$36 { r }^{ 2 } -24r-5$$

Factor

$$\left(6r-5\right)\left(6r+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 36r^{2}+ar+br-5. To find a and b, set up a system to be solved.

$$a+b=-24 ab=36\left(-5\right)=-180$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -180.

$$1,-180 2,-90 3,-60 4,-45 5,-36 6,-30 9,-20 10,-18 12,-15$$

Calculate the sum for each pair.

$$1-180=-179 2-90=-88 3-60=-57 4-45=-41 5-36=-31 6-30=-24 9-20=-11 10-18=-8 12-15=-3$$

The solution is the pair that gives sum -24.

$$a=-30 b=6$$

Rewrite 36r^{2}-24r-5 as \left(36r^{2}-30r\right)+\left(6r-5\right).

$$\left(36r^{2}-30r\right)+\left(6r-5\right)$$

Factor out 6r in 36r^{2}-30r.

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

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

$$\left(6r-5\right)\left(6r+1\right)$$

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.

$$36r^{2}-24r-5=0$$

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.

$$r=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 36\left(-5\right)}}{2\times 36}$$

Square -24.

$$r=\frac{-\left(-24\right)±\sqrt{576-4\times 36\left(-5\right)}}{2\times 36}$$

Multiply -4 times 36.

$$r=\frac{-\left(-24\right)±\sqrt{576-144\left(-5\right)}}{2\times 36}$$

Multiply -144 times -5.

$$r=\frac{-\left(-24\right)±\sqrt{576+720}}{2\times 36}$$

Add 576 to 720.

$$r=\frac{-\left(-24\right)±\sqrt{1296}}{2\times 36}$$

Take the square root of 1296.

$$r=\frac{-\left(-24\right)±36}{2\times 36}$$

The opposite of -24 is 24.

$$r=\frac{24±36}{2\times 36}$$

Multiply 2 times 36.

$$r=\frac{24±36}{72}$$

Now solve the equation r=\frac{24±36}{72} when ± is plus. Add 24 to 36.

$$r=\frac{60}{72}$$

Reduce the fraction \frac{60}{72} to lowest terms by extracting and canceling out 12.

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

Now solve the equation r=\frac{24±36}{72} when ± is minus. Subtract 36 from 24.

$$r=-\frac{12}{72}$$

Reduce the fraction \frac{-12}{72} to lowest terms by extracting and canceling out 12.

$$r=-\frac{1}{6}$$

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{1}{6} for x_{2}.

$$36r^{2}-24r-5=36\left(r-\frac{5}{6}\right)\left(r-\left(-\frac{1}{6}\right)\right)$$

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

$$36r^{2}-24r-5=36\left(r-\frac{5}{6}\right)\left(r+\frac{1}{6}\right)$$

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

$$36r^{2}-24r-5=36\times \frac{6r-5}{6}\left(r+\frac{1}{6}\right)$$

Add \frac{1}{6} to r by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$36r^{2}-24r-5=36\times \frac{6r-5}{6}\times \frac{6r+1}{6}$$

Multiply \frac{6r-5}{6} times \frac{6r+1}{6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$36r^{2}-24r-5=36\times \frac{\left(6r-5\right)\left(6r+1\right)}{6\times 6}$$

Multiply 6 times 6.

$$36r^{2}-24r-5=36\times \frac{\left(6r-5\right)\left(6r+1\right)}{36}$$

Cancel out 36, the greatest common factor in 36 and 36.

$$36r^{2}-24r-5=\left(6r-5\right)\left(6r+1\right)$$

Evaluate

$$\left(6r-5\right)\left(6r+1\right)$$