Solve 5y^2-9y-18

$$5 y ^ { 2 } - 9 y - 18$$

Factor

$$\left(y-3\right)\left(5y+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5y^{2}+ay+by-18. To find a and b, set up a system to be solved.

$$a+b=-9 ab=5\left(-18\right)=-90$$

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 -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -9.

$$a=-15 b=6$$

Rewrite 5y^{2}-9y-18 as \left(5y^{2}-15y\right)+\left(6y-18\right).

$$\left(5y^{2}-15y\right)+\left(6y-18\right)$$

Factor out 5y in the first and 6 in the second group.

$$5y\left(y-3\right)+6\left(y-3\right)$$

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

$$\left(y-3\right)\left(5y+6\right)$$

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.

$$5y^{2}-9y-18=0$$

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.

$$y=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 5\left(-18\right)}}{2\times 5}$$

Square -9.

$$y=\frac{-\left(-9\right)±\sqrt{81-4\times 5\left(-18\right)}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-\left(-9\right)±\sqrt{81-20\left(-18\right)}}{2\times 5}$$

Multiply -20 times -18.

$$y=\frac{-\left(-9\right)±\sqrt{81+360}}{2\times 5}$$

Add 81 to 360.

$$y=\frac{-\left(-9\right)±\sqrt{441}}{2\times 5}$$

Take the square root of 441.

$$y=\frac{-\left(-9\right)±21}{2\times 5}$$

The opposite of -9 is 9.

$$y=\frac{9±21}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{9±21}{10}$$

Now solve the equation y=\frac{9±21}{10} when ± is plus. Add 9 to 21.

$$y=\frac{30}{10}$$

Divide 30 by 10.

$$y=3$$

Now solve the equation y=\frac{9±21}{10} when ± is minus. Subtract 21 from 9.

$$y=-\frac{12}{10}$$

Reduce the fraction \frac{-12}{10} to lowest terms by extracting and canceling out 2.

$$y=-\frac{6}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and -\frac{6}{5} for x_{2}.

$$5y^{2}-9y-18=5\left(y-3\right)\left(y-\left(-\frac{6}{5}\right)\right)$$

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

$$5y^{2}-9y-18=5\left(y-3\right)\left(y+\frac{6}{5}\right)$$

Add \frac{6}{5} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5y^{2}-9y-18=5\left(y-3\right)\times \frac{5y+6}{5}$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5y^{2}-9y-18=\left(y-3\right)\left(5y+6\right)$$

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 5

$$x ^ 2 -\frac{9}{5}x -\frac{18}{5} = 0$$

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

$$r + s = \frac{9}{5} rs = -\frac{18}{5}$$

Two numbers r and s sum up to \frac{9}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{5} = \frac{9}{10}. 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>

$$r = \frac{9}{10} - u s = \frac{9}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{18}{5}

$$(\frac{9}{10} - u) (\frac{9}{10} + u) = -\frac{18}{5}$$

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

$$\frac{81}{100} - u^2 = -\frac{18}{5}$$

Simplify the expression by subtracting \frac{81}{100} on both sides

$$-u^2 = -\frac{18}{5}-\frac{81}{100} = -\frac{441}{100}$$

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

$$u^2 = \frac{441}{100} u = \pm\sqrt{\frac{441}{100}} = \pm \frac{21}{10}$$

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

$$r =\frac{9}{10} - \frac{21}{10} = -1.200 s = \frac{9}{10} + \frac{21}{10} = 3$$

Evaluate

$$\left(y-3\right)\left(5y+6\right)$$