Solve 9y^2-104y-48

$$9 y ^ { 2 } - 104 y - 48$$

Factor

$$\left(y-12\right)\left(9y+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9y^{2}+ay+by-48. To find a and b, set up a system to be solved.

$$a+b=-104 ab=9\left(-48\right)=-432$$

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 -432.

$$1,-432 2,-216 3,-144 4,-108 6,-72 8,-54 9,-48 12,-36 16,-27 18,-24$$

Calculate the sum for each pair.

$$1-432=-431 2-216=-214 3-144=-141 4-108=-104 6-72=-66 8-54=-46 9-48=-39 12-36=-24 16-27=-11 18-24=-6$$

The solution is the pair that gives sum -104.

$$a=-108 b=4$$

Rewrite 9y^{2}-104y-48 as \left(9y^{2}-108y\right)+\left(4y-48\right).

$$\left(9y^{2}-108y\right)+\left(4y-48\right)$$

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

$$9y\left(y-12\right)+4\left(y-12\right)$$

Factor out common term y-12 by using distributive property.

$$\left(y-12\right)\left(9y+4\right)$$

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.

$$9y^{2}-104y-48=0$$

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.

$$y=\frac{-\left(-104\right)±\sqrt{\left(-104\right)^{2}-4\times 9\left(-48\right)}}{2\times 9}$$

Square -104.

$$y=\frac{-\left(-104\right)±\sqrt{10816-4\times 9\left(-48\right)}}{2\times 9}$$

Multiply -4 times 9.

$$y=\frac{-\left(-104\right)±\sqrt{10816-36\left(-48\right)}}{2\times 9}$$

Multiply -36 times -48.

$$y=\frac{-\left(-104\right)±\sqrt{10816+1728}}{2\times 9}$$

Add 10816 to 1728.

$$y=\frac{-\left(-104\right)±\sqrt{12544}}{2\times 9}$$

Take the square root of 12544.

$$y=\frac{-\left(-104\right)±112}{2\times 9}$$

The opposite of -104 is 104.

$$y=\frac{104±112}{2\times 9}$$

Multiply 2 times 9.

$$y=\frac{104±112}{18}$$

Now solve the equation y=\frac{104±112}{18} when ± is plus. Add 104 to 112.

$$y=\frac{216}{18}$$

Divide 216 by 18.

$$y=12$$

Now solve the equation y=\frac{104±112}{18} when ± is minus. Subtract 112 from 104.

$$y=-\frac{8}{18}$$

Reduce the fraction \frac{-8}{18} to lowest terms by extracting and canceling out 2.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 12 for x_{1} and -\frac{4}{9} for x_{2}.

$$9y^{2}-104y-48=9\left(y-12\right)\left(y-\left(-\frac{4}{9}\right)\right)$$

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

$$9y^{2}-104y-48=9\left(y-12\right)\left(y+\frac{4}{9}\right)$$

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

$$9y^{2}-104y-48=9\left(y-12\right)\times \frac{9y+4}{9}$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9y^{2}-104y-48=\left(y-12\right)\left(9y+4\right)$$

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 9

$$x ^ 2 -\frac{104}{9}x -\frac{16}{3} = 0$$

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

$$r + s = \frac{104}{9} rs = -\frac{16}{3}$$

Two numbers r and s sum up to \frac{104}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{104}{9} = \frac{52}{9}. 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>

$$r = \frac{52}{9} - u s = \frac{52}{9} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{16}{3}

$$(\frac{52}{9} - u) (\frac{52}{9} + u) = -\frac{16}{3}$$

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

$$\frac{2704}{81} - u^2 = -\frac{16}{3}$$

Simplify the expression by subtracting \frac{2704}{81} on both sides

$$-u^2 = -\frac{16}{3}-\frac{2704}{81} = -\frac{3136}{81}$$

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

$$u^2 = \frac{3136}{81} u = \pm\sqrt{\frac{3136}{81}} = \pm \frac{56}{9}$$

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

$$r =\frac{52}{9} - \frac{56}{9} = -0.444 s = \frac{52}{9} + \frac{56}{9} = 12$$

Evaluate

$$\left(y-12\right)\left(9y+4\right)$$