Solve 2y^2-9y-18

$$2 y ^ { 2 } - 9 y - 18$$

Factor

$$\left(y-6\right)\left(2y+3\right)$$

Solution Steps

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

$$a+b=-9 ab=2\left(-18\right)=-36$$

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 -36.

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

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -9.

$$a=-12 b=3$$

Rewrite 2y^{2}-9y-18 as \left(2y^{2}-12y\right)+\left(3y-18\right).

$$\left(2y^{2}-12y\right)+\left(3y-18\right)$$

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

$$2y\left(y-6\right)+3\left(y-6\right)$$

Factor out common term y-6 by using distributive property.

$$\left(y-6\right)\left(2y+3\right)$$

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.

$$2y^{2}-9y-18=0$$

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.

$$y=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\left(-18\right)}}{2\times 2}$$

Square -9.

$$y=\frac{-\left(-9\right)±\sqrt{81-4\times 2\left(-18\right)}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-\left(-9\right)±\sqrt{81-8\left(-18\right)}}{2\times 2}$$

Multiply -8 times -18.

$$y=\frac{-\left(-9\right)±\sqrt{81+144}}{2\times 2}$$

Add 81 to 144.

$$y=\frac{-\left(-9\right)±\sqrt{225}}{2\times 2}$$

Take the square root of 225.

$$y=\frac{-\left(-9\right)±15}{2\times 2}$$

The opposite of -9 is 9.

$$y=\frac{9±15}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{9±15}{4}$$

Now solve the equation y=\frac{9±15}{4} when ± is plus. Add 9 to 15.

$$y=\frac{24}{4}$$

Divide 24 by 4.

$$y=6$$

Now solve the equation y=\frac{9±15}{4} when ± is minus. Subtract 15 from 9.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and -\frac{3}{2} for x_{2}.

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

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

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

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

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

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

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

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

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

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

$$r + s = \frac{9}{2} rs = -9$$

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

$$r = \frac{9}{4} - u s = \frac{9}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -9

$$(\frac{9}{4} - u) (\frac{9}{4} + u) = -9$$

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

$$\frac{81}{16} - u^2 = -9$$

Simplify the expression by subtracting \frac{81}{16} on both sides

$$-u^2 = -9-\frac{81}{16} = -\frac{225}{16}$$

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

$$u^2 = \frac{225}{16} u = \pm\sqrt{\frac{225}{16}} = \pm \frac{15}{4}$$

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

$$r =\frac{9}{4} - \frac{15}{4} = -1.500 s = \frac{9}{4} + \frac{15}{4} = 6$$

Evaluate

$$\left(y-6\right)\left(2y+3\right)$$