Solve 18y^2-23y-6

$$18 y ^ { 2 } - 23 y - 6$$

Factor

$$\left(2y-3\right)\left(9y+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 18y^{2}+ay+by-6. To find a and b, set up a system to be solved.

$$a+b=-23 ab=18\left(-6\right)=-108$$

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 -108.

$$1,-108 2,-54 3,-36 4,-27 6,-18 9,-12$$

Calculate the sum for each pair.

$$1-108=-107 2-54=-52 3-36=-33 4-27=-23 6-18=-12 9-12=-3$$

The solution is the pair that gives sum -23.

$$a=-27 b=4$$

Rewrite 18y^{2}-23y-6 as \left(18y^{2}-27y\right)+\left(4y-6\right).

$$\left(18y^{2}-27y\right)+\left(4y-6\right)$$

Factor out 9y in the first and 2 in the second group.

$$9y\left(2y-3\right)+2\left(2y-3\right)$$

Factor out common term 2y-3 by using distributive property.

$$\left(2y-3\right)\left(9y+2\right)$$

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.

$$18y^{2}-23y-6=0$$

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.

$$y=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 18\left(-6\right)}}{2\times 18}$$

Square -23.

$$y=\frac{-\left(-23\right)±\sqrt{529-4\times 18\left(-6\right)}}{2\times 18}$$

Multiply -4 times 18.

$$y=\frac{-\left(-23\right)±\sqrt{529-72\left(-6\right)}}{2\times 18}$$

Multiply -72 times -6.

$$y=\frac{-\left(-23\right)±\sqrt{529+432}}{2\times 18}$$

Add 529 to 432.

$$y=\frac{-\left(-23\right)±\sqrt{961}}{2\times 18}$$

Take the square root of 961.

$$y=\frac{-\left(-23\right)±31}{2\times 18}$$

The opposite of -23 is 23.

$$y=\frac{23±31}{2\times 18}$$

Multiply 2 times 18.

$$y=\frac{23±31}{36}$$

Now solve the equation y=\frac{23±31}{36} when ± is plus. Add 23 to 31.

$$y=\frac{54}{36}$$

Reduce the fraction \frac{54}{36} to lowest terms by extracting and canceling out 18.

$$y=\frac{3}{2}$$

Now solve the equation y=\frac{23±31}{36} when ± is minus. Subtract 31 from 23.

$$y=-\frac{8}{36}$$

Reduce the fraction \frac{-8}{36} to lowest terms by extracting and canceling out 4.

$$y=-\frac{2}{9}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and -\frac{2}{9} for x_{2}.

$$18y^{2}-23y-6=18\left(y-\frac{3}{2}\right)\left(y-\left(-\frac{2}{9}\right)\right)$$

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

$$18y^{2}-23y-6=18\left(y-\frac{3}{2}\right)\left(y+\frac{2}{9}\right)$$

Subtract \frac{3}{2} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18y^{2}-23y-6=18\times \frac{2y-3}{2}\left(y+\frac{2}{9}\right)$$

Add \frac{2}{9} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18y^{2}-23y-6=18\times \frac{2y-3}{2}\times \frac{9y+2}{9}$$

Multiply \frac{2y-3}{2} times \frac{9y+2}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18y^{2}-23y-6=18\times \frac{\left(2y-3\right)\left(9y+2\right)}{2\times 9}$$

Multiply 2 times 9.

$$18y^{2}-23y-6=18\times \frac{\left(2y-3\right)\left(9y+2\right)}{18}$$

Cancel out 18, the greatest common factor in 18 and 18.

$$18y^{2}-23y-6=\left(2y-3\right)\left(9y+2\right)$$

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 18

$$x ^ 2 -\frac{23}{18}x -\frac{1}{3} = 0$$

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

$$r + s = \frac{23}{18} rs = -\frac{1}{3}$$

Two numbers r and s sum up to \frac{23}{18} exactly when the average of the two numbers is \frac{1}{2}*\frac{23}{18} = \frac{23}{36}. 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>

$$r = \frac{23}{36} - u s = \frac{23}{36} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{3}

$$(\frac{23}{36} - u) (\frac{23}{36} + u) = -\frac{1}{3}$$

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

$$\frac{529}{1296} - u^2 = -\frac{1}{3}$$

Simplify the expression by subtracting \frac{529}{1296} on both sides

$$-u^2 = -\frac{1}{3}-\frac{529}{1296} = -\frac{961}{1296}$$

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

$$u^2 = \frac{961}{1296} u = \pm\sqrt{\frac{961}{1296}} = \pm \frac{31}{36}$$

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

$$r =\frac{23}{36} - \frac{31}{36} = -0.222 s = \frac{23}{36} + \frac{31}{36} = 1.500$$

Evaluate

$$\left(2y-3\right)\left(9y+2\right)$$