Solve 18y^2+3y-45

$$18 y ^ { 2 } + 3 y - 45$$

Factor

$$3\left(2y-3\right)\left(3y+5\right)$$

Solution Steps

Factor out 3.

$$3\left(6y^{2}+y-15\right)$$

Consider 6y^{2}+y-15. Factor the expression by grouping. First, the expression needs to be rewritten as 6y^{2}+ay+by-15. To find a and b, set up a system to be solved.

$$a+b=1 ab=6\left(-15\right)=-90$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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 1.

$$a=-9 b=10$$

Rewrite 6y^{2}+y-15 as \left(6y^{2}-9y\right)+\left(10y-15\right).

$$\left(6y^{2}-9y\right)+\left(10y-15\right)$$

Factor out 3y in the first and 5 in the second group.

$$3y\left(2y-3\right)+5\left(2y-3\right)$$

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

$$\left(2y-3\right)\left(3y+5\right)$$

Rewrite the complete factored expression.

$$3\left(2y-3\right)\left(3y+5\right)$$

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.

$$18y^{2}+3y-45=0$$

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.

$$y=\frac{-3±\sqrt{3^{2}-4\times 18\left(-45\right)}}{2\times 18}$$

Square 3.

$$y=\frac{-3±\sqrt{9-4\times 18\left(-45\right)}}{2\times 18}$$

Multiply -4 times 18.

$$y=\frac{-3±\sqrt{9-72\left(-45\right)}}{2\times 18}$$

Multiply -72 times -45.

$$y=\frac{-3±\sqrt{9+3240}}{2\times 18}$$

Add 9 to 3240.

$$y=\frac{-3±\sqrt{3249}}{2\times 18}$$

Take the square root of 3249.

$$y=\frac{-3±57}{2\times 18}$$

Multiply 2 times 18.

$$y=\frac{-3±57}{36}$$

Now solve the equation y=\frac{-3±57}{36} when ± is plus. Add -3 to 57.

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

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{-3±57}{36} when ± is minus. Subtract 57 from -3.

$$y=-\frac{60}{36}$$

Reduce the fraction \frac{-60}{36} to lowest terms by extracting and canceling out 12.

$$y=-\frac{5}{3}$$

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{5}{3} for x_{2}.

$$18y^{2}+3y-45=18\left(y-\frac{3}{2}\right)\left(y-\left(-\frac{5}{3}\right)\right)$$

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

$$18y^{2}+3y-45=18\left(y-\frac{3}{2}\right)\left(y+\frac{5}{3}\right)$$

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

$$18y^{2}+3y-45=18\times \frac{2y-3}{2}\left(y+\frac{5}{3}\right)$$

Add \frac{5}{3} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18y^{2}+3y-45=18\times \frac{2y-3}{2}\times \frac{3y+5}{3}$$

Multiply \frac{2y-3}{2} times \frac{3y+5}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18y^{2}+3y-45=18\times \frac{\left(2y-3\right)\left(3y+5\right)}{2\times 3}$$

Multiply 2 times 3.

$$18y^{2}+3y-45=18\times \frac{\left(2y-3\right)\left(3y+5\right)}{6}$$

Cancel out 6, the greatest common factor in 18 and 6.

$$18y^{2}+3y-45=3\left(2y-3\right)\left(3y+5\right)$$

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

$$x ^ 2 +\frac{1}{6}x -\frac{5}{2} = 0$$

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

$$r + s = -\frac{1}{6} rs = -\frac{5}{2}$$

Two numbers r and s sum up to -\frac{1}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{6} = -\frac{1}{12}. 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>

$$r = -\frac{1}{12} - u s = -\frac{1}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{2}

$$(-\frac{1}{12} - u) (-\frac{1}{12} + u) = -\frac{5}{2}$$

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

$$\frac{1}{144} - u^2 = -\frac{5}{2}$$

Simplify the expression by subtracting \frac{1}{144} on both sides

$$-u^2 = -\frac{5}{2}-\frac{1}{144} = -\frac{361}{144}$$

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

$$u^2 = \frac{361}{144} u = \pm\sqrt{\frac{361}{144}} = \pm \frac{19}{12}$$

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

$$r =-\frac{1}{12} - \frac{19}{12} = -1.667 s = -\frac{1}{12} + \frac{19}{12} = 1.500$$

Evaluate

$$3\left(6y^{2}+y-15\right)$$