Solve 8r^2-2r-45

$$8 r ^ { 2 } - 2 r - 45$$

Factor

$$\left(2r-5\right)\left(4r+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8r^{2}+ar+br-45. To find a and b, set up a system to be solved.

$$a+b=-2 ab=8\left(-45\right)=-360$$

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 -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 -2.

$$a=-20 b=18$$

Rewrite 8r^{2}-2r-45 as \left(8r^{2}-20r\right)+\left(18r-45\right).

$$\left(8r^{2}-20r\right)+\left(18r-45\right)$$

Factor out 4r in the first and 9 in the second group.

$$4r\left(2r-5\right)+9\left(2r-5\right)$$

Factor out common term 2r-5 by using distributive property.

$$\left(2r-5\right)\left(4r+9\right)$$

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.

$$8r^{2}-2r-45=0$$

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.

$$r=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 8\left(-45\right)}}{2\times 8}$$

Square -2.

$$r=\frac{-\left(-2\right)±\sqrt{4-4\times 8\left(-45\right)}}{2\times 8}$$

Multiply -4 times 8.

$$r=\frac{-\left(-2\right)±\sqrt{4-32\left(-45\right)}}{2\times 8}$$

Multiply -32 times -45.

$$r=\frac{-\left(-2\right)±\sqrt{4+1440}}{2\times 8}$$

Add 4 to 1440.

$$r=\frac{-\left(-2\right)±\sqrt{1444}}{2\times 8}$$

Take the square root of 1444.

$$r=\frac{-\left(-2\right)±38}{2\times 8}$$

The opposite of -2 is 2.

$$r=\frac{2±38}{2\times 8}$$

Multiply 2 times 8.

$$r=\frac{2±38}{16}$$

Now solve the equation r=\frac{2±38}{16} when ± is plus. Add 2 to 38.

$$r=\frac{40}{16}$$

Reduce the fraction \frac{40}{16} to lowest terms by extracting and canceling out 8.

$$r=\frac{5}{2}$$

Now solve the equation r=\frac{2±38}{16} when ± is minus. Subtract 38 from 2.

$$r=-\frac{36}{16}$$

Reduce the fraction \frac{-36}{16} to lowest terms by extracting and canceling out 4.

$$r=-\frac{9}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and -\frac{9}{4} for x_{2}.

$$8r^{2}-2r-45=8\left(r-\frac{5}{2}\right)\left(r-\left(-\frac{9}{4}\right)\right)$$

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

$$8r^{2}-2r-45=8\left(r-\frac{5}{2}\right)\left(r+\frac{9}{4}\right)$$

Subtract \frac{5}{2} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8r^{2}-2r-45=8\times \frac{2r-5}{2}\left(r+\frac{9}{4}\right)$$

Add \frac{9}{4} to r by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$8r^{2}-2r-45=8\times \frac{2r-5}{2}\times \frac{4r+9}{4}$$

Multiply \frac{2r-5}{2} times \frac{4r+9}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$8r^{2}-2r-45=8\times \frac{\left(2r-5\right)\left(4r+9\right)}{2\times 4}$$

Multiply 2 times 4.

$$8r^{2}-2r-45=8\times \frac{\left(2r-5\right)\left(4r+9\right)}{8}$$

Cancel out 8, the greatest common factor in 8 and 8.

$$8r^{2}-2r-45=\left(2r-5\right)\left(4r+9\right)$$

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 8

$$x ^ 2 -\frac{1}{4}x -\frac{45}{8} = 0$$

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

$$r + s = \frac{1}{4} rs = -\frac{45}{8}$$

Two numbers r and s sum up to \frac{1}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{4} = \frac{1}{8}. 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>

$$r = \frac{1}{8} - u s = \frac{1}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{45}{8}

$$(\frac{1}{8} - u) (\frac{1}{8} + u) = -\frac{45}{8}$$

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

$$\frac{1}{64} - u^2 = -\frac{45}{8}$$

Simplify the expression by subtracting \frac{1}{64} on both sides

$$-u^2 = -\frac{45}{8}-\frac{1}{64} = -\frac{361}{64}$$

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

$$u^2 = \frac{361}{64} u = \pm\sqrt{\frac{361}{64}} = \pm \frac{19}{8}$$

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

$$r =\frac{1}{8} - \frac{19}{8} = -2.250 s = \frac{1}{8} + \frac{19}{8} = 2.500$$

Evaluate

$$\left(2r-5\right)\left(4r+9\right)$$