Solve 60r^2-42r-36

$$60 r ^ { 2 } - 42 r - 36$$

Factor

$$6\left(5r-6\right)\left(2r+1\right)$$

Solution Steps

Factor out 6.

$$6\left(10r^{2}-7r-6\right)$$

Consider 10r^{2}-7r-6. Factor the expression by grouping. First, the expression needs to be rewritten as 10r^{2}+ar+br-6. To find a and b, set up a system to be solved.

$$a+b=-7 ab=10\left(-6\right)=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -7.

$$a=-12 b=5$$

Rewrite 10r^{2}-7r-6 as \left(10r^{2}-12r\right)+\left(5r-6\right).

$$\left(10r^{2}-12r\right)+\left(5r-6\right)$$

Factor out 2r in 10r^{2}-12r.

$$2r\left(5r-6\right)+5r-6$$

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

$$\left(5r-6\right)\left(2r+1\right)$$

Rewrite the complete factored expression.

$$6\left(5r-6\right)\left(2r+1\right)$$

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.

$$60r^{2}-42r-36=0$$

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.

$$r=\frac{-\left(-42\right)±\sqrt{\left(-42\right)^{2}-4\times 60\left(-36\right)}}{2\times 60}$$

Square -42.

$$r=\frac{-\left(-42\right)±\sqrt{1764-4\times 60\left(-36\right)}}{2\times 60}$$

Multiply -4 times 60.

$$r=\frac{-\left(-42\right)±\sqrt{1764-240\left(-36\right)}}{2\times 60}$$

Multiply -240 times -36.

$$r=\frac{-\left(-42\right)±\sqrt{1764+8640}}{2\times 60}$$

Add 1764 to 8640.

$$r=\frac{-\left(-42\right)±\sqrt{10404}}{2\times 60}$$

Take the square root of 10404.

$$r=\frac{-\left(-42\right)±102}{2\times 60}$$

The opposite of -42 is 42.

$$r=\frac{42±102}{2\times 60}$$

Multiply 2 times 60.

$$r=\frac{42±102}{120}$$

Now solve the equation r=\frac{42±102}{120} when ± is plus. Add 42 to 102.

$$r=\frac{144}{120}$$

Reduce the fraction \frac{144}{120} to lowest terms by extracting and canceling out 24.

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

Now solve the equation r=\frac{42±102}{120} when ± is minus. Subtract 102 from 42.

$$r=-\frac{60}{120}$$

Reduce the fraction \frac{-60}{120} to lowest terms by extracting and canceling out 60.

$$r=-\frac{1}{2}$$

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

$$60r^{2}-42r-36=60\left(r-\frac{6}{5}\right)\left(r-\left(-\frac{1}{2}\right)\right)$$

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

$$60r^{2}-42r-36=60\left(r-\frac{6}{5}\right)\left(r+\frac{1}{2}\right)$$

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

$$60r^{2}-42r-36=60\times \frac{5r-6}{5}\left(r+\frac{1}{2}\right)$$

Add \frac{1}{2} to r by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$60r^{2}-42r-36=60\times \frac{5r-6}{5}\times \frac{2r+1}{2}$$

Multiply \frac{5r-6}{5} times \frac{2r+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$60r^{2}-42r-36=60\times \frac{\left(5r-6\right)\left(2r+1\right)}{5\times 2}$$

Multiply 5 times 2.

$$60r^{2}-42r-36=60\times \frac{\left(5r-6\right)\left(2r+1\right)}{10}$$

Cancel out 10, the greatest common factor in 60 and 10.

$$60r^{2}-42r-36=6\left(5r-6\right)\left(2r+1\right)$$

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 60

$$x ^ 2 -\frac{7}{10}x -\frac{3}{5} = 0$$

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

$$r + s = \frac{7}{10} rs = -\frac{3}{5}$$

Two numbers r and s sum up to \frac{7}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{10} = \frac{7}{20}. 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>

$$r = \frac{7}{20} - u s = \frac{7}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{5}

$$(\frac{7}{20} - u) (\frac{7}{20} + u) = -\frac{3}{5}$$

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

$$\frac{49}{400} - u^2 = -\frac{3}{5}$$

Simplify the expression by subtracting \frac{49}{400} on both sides

$$-u^2 = -\frac{3}{5}-\frac{49}{400} = -\frac{289}{400}$$

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

$$u^2 = \frac{289}{400} u = \pm\sqrt{\frac{289}{400}} = \pm \frac{17}{20}$$

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

$$r =\frac{7}{20} - \frac{17}{20} = -0.500 s = \frac{7}{20} + \frac{17}{20} = 1.200$$

Evaluate

$$60r^{2}-42r-36$$