Solve 2r^2-13r-24

$$2 r ^ { 2 } - 13 r - 24$$

Factor

$$\left(r-8\right)\left(2r+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2r^{2}+ar+br-24. To find a and b, set up a system to be solved.

$$a+b=-13 ab=2\left(-24\right)=-48$$

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 -48.

$$1,-48 2,-24 3,-16 4,-12 6,-8$$

Calculate the sum for each pair.

$$1-48=-47 2-24=-22 3-16=-13 4-12=-8 6-8=-2$$

The solution is the pair that gives sum -13.

$$a=-16 b=3$$

Rewrite 2r^{2}-13r-24 as \left(2r^{2}-16r\right)+\left(3r-24\right).

$$\left(2r^{2}-16r\right)+\left(3r-24\right)$$

Factor out 2r in the first and 3 in the second group.

$$2r\left(r-8\right)+3\left(r-8\right)$$

Factor out common term r-8 by using distributive property.

$$\left(r-8\right)\left(2r+3\right)$$

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.

$$2r^{2}-13r-24=0$$

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.

$$r=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 2\left(-24\right)}}{2\times 2}$$

Square -13.

$$r=\frac{-\left(-13\right)±\sqrt{169-4\times 2\left(-24\right)}}{2\times 2}$$

Multiply -4 times 2.

$$r=\frac{-\left(-13\right)±\sqrt{169-8\left(-24\right)}}{2\times 2}$$

Multiply -8 times -24.

$$r=\frac{-\left(-13\right)±\sqrt{169+192}}{2\times 2}$$

Add 169 to 192.

$$r=\frac{-\left(-13\right)±\sqrt{361}}{2\times 2}$$

Take the square root of 361.

$$r=\frac{-\left(-13\right)±19}{2\times 2}$$

The opposite of -13 is 13.

$$r=\frac{13±19}{2\times 2}$$

Multiply 2 times 2.

$$r=\frac{13±19}{4}$$

Now solve the equation r=\frac{13±19}{4} when ± is plus. Add 13 to 19.

$$r=\frac{32}{4}$$

Divide 32 by 4.

$$r=8$$

Now solve the equation r=\frac{13±19}{4} when ± is minus. Subtract 19 from 13.

$$r=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$r=-\frac{3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and -\frac{3}{2} for x_{2}.

$$2r^{2}-13r-24=2\left(r-8\right)\left(r-\left(-\frac{3}{2}\right)\right)$$

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

$$2r^{2}-13r-24=2\left(r-8\right)\left(r+\frac{3}{2}\right)$$

Add \frac{3}{2} to r by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$2r^{2}-13r-24=2\left(r-8\right)\times \frac{2r+3}{2}$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2r^{2}-13r-24=\left(r-8\right)\left(2r+3\right)$$

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 2

$$x ^ 2 -\frac{13}{2}x -12 = 0$$

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

$$r + s = \frac{13}{2} rs = -12$$

Two numbers r and s sum up to \frac{13}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{2} = \frac{13}{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>

$$r = \frac{13}{4} - u s = \frac{13}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -12

$$(\frac{13}{4} - u) (\frac{13}{4} + u) = -12$$

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

$$\frac{169}{16} - u^2 = -12$$

Simplify the expression by subtracting \frac{169}{16} on both sides

$$-u^2 = -12-\frac{169}{16} = -\frac{361}{16}$$

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

$$u^2 = \frac{361}{16} u = \pm\sqrt{\frac{361}{16}} = \pm \frac{19}{4}$$

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

$$r =\frac{13}{4} - \frac{19}{4} = -1.500 s = \frac{13}{4} + \frac{19}{4} = 8$$

Evaluate

$$\left(r-8\right)\left(2r+3\right)$$