Solve 8r^2+22r-40

$$8 r ^ { 2 } + 22 r - 40$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(4r^{2}+11r-20\right)$$

Consider 4r^{2}+11r-20. Factor the expression by grouping. First, the expression needs to be rewritten as 4r^{2}+ar+br-20. To find a and b, set up a system to be solved.

$$a+b=11 ab=4\left(-20\right)=-80$$

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 -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 11.

$$a=-5 b=16$$

Rewrite 4r^{2}+11r-20 as \left(4r^{2}-5r\right)+\left(16r-20\right).

$$\left(4r^{2}-5r\right)+\left(16r-20\right)$$

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

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

Factor out common term 4r-5 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$8r^{2}+22r-40=0$$

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.

$$r=\frac{-22±\sqrt{22^{2}-4\times 8\left(-40\right)}}{2\times 8}$$

Square 22.

$$r=\frac{-22±\sqrt{484-4\times 8\left(-40\right)}}{2\times 8}$$

Multiply -4 times 8.

$$r=\frac{-22±\sqrt{484-32\left(-40\right)}}{2\times 8}$$

Multiply -32 times -40.

$$r=\frac{-22±\sqrt{484+1280}}{2\times 8}$$

Add 484 to 1280.

$$r=\frac{-22±\sqrt{1764}}{2\times 8}$$

Take the square root of 1764.

$$r=\frac{-22±42}{2\times 8}$$

Multiply 2 times 8.

$$r=\frac{-22±42}{16}$$

Now solve the equation r=\frac{-22±42}{16} when ± is plus. Add -22 to 42.

$$r=\frac{20}{16}$$

Reduce the fraction \frac{20}{16} to lowest terms by extracting and canceling out 4.

$$r=\frac{5}{4}$$

Now solve the equation r=\frac{-22±42}{16} when ± is minus. Subtract 42 from -22.

$$r=-\frac{64}{16}$$

Divide -64 by 16.

$$r=-4$$

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

$$8r^{2}+22r-40=8\left(r-\frac{5}{4}\right)\left(r-\left(-4\right)\right)$$

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

$$8r^{2}+22r-40=8\left(r-\frac{5}{4}\right)\left(r+4\right)$$

Subtract \frac{5}{4} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8r^{2}+22r-40=8\times \frac{4r-5}{4}\left(r+4\right)$$

Cancel out 4, the greatest common factor in 8 and 4.

$$8r^{2}+22r-40=2\left(4r-5\right)\left(r+4\right)$$

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 8

$$x ^ 2 +\frac{11}{4}x -5 = 0$$

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

$$r + s = -\frac{11}{4} rs = -5$$

Two numbers r and s sum up to -\frac{11}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{4} = -\frac{11}{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>

$$r = -\frac{11}{8} - u s = -\frac{11}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -5

$$(-\frac{11}{8} - u) (-\frac{11}{8} + u) = -5$$

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

$$\frac{121}{64} - u^2 = -5$$

Simplify the expression by subtracting \frac{121}{64} on both sides

$$-u^2 = -5-\frac{121}{64} = -\frac{441}{64}$$

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

$$u^2 = \frac{441}{64} u = \pm\sqrt{\frac{441}{64}} = \pm \frac{21}{8}$$

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

$$r =-\frac{11}{8} - \frac{21}{8} = -4 s = -\frac{11}{8} + \frac{21}{8} = 1.250$$

Evaluate

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