Solve 9y^2+75y-54

$$9 y ^ { 2 } + 75 y - 54$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(3y^{2}+25y-18\right)$$

Consider 3y^{2}+25y-18. Factor the expression by grouping. First, the expression needs to be rewritten as 3y^{2}+ay+by-18. To find a and b, set up a system to be solved.

$$a+b=25 ab=3\left(-18\right)=-54$$

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 -54.

$$-1,54 -2,27 -3,18 -6,9$$

Calculate the sum for each pair.

$$-1+54=53 -2+27=25 -3+18=15 -6+9=3$$

The solution is the pair that gives sum 25.

$$a=-2 b=27$$

Rewrite 3y^{2}+25y-18 as \left(3y^{2}-2y\right)+\left(27y-18\right).

$$\left(3y^{2}-2y\right)+\left(27y-18\right)$$

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$9y^{2}+75y-54=0$$

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.

$$y=\frac{-75±\sqrt{75^{2}-4\times 9\left(-54\right)}}{2\times 9}$$

Square 75.

$$y=\frac{-75±\sqrt{5625-4\times 9\left(-54\right)}}{2\times 9}$$

Multiply -4 times 9.

$$y=\frac{-75±\sqrt{5625-36\left(-54\right)}}{2\times 9}$$

Multiply -36 times -54.

$$y=\frac{-75±\sqrt{5625+1944}}{2\times 9}$$

Add 5625 to 1944.

$$y=\frac{-75±\sqrt{7569}}{2\times 9}$$

Take the square root of 7569.

$$y=\frac{-75±87}{2\times 9}$$

Multiply 2 times 9.

$$y=\frac{-75±87}{18}$$

Now solve the equation y=\frac{-75±87}{18} when ± is plus. Add -75 to 87.

$$y=\frac{12}{18}$$

Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation y=\frac{-75±87}{18} when ± is minus. Subtract 87 from -75.

$$y=-\frac{162}{18}$$

Divide -162 by 18.

$$y=-9$$

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

$$9y^{2}+75y-54=9\left(y-\frac{2}{3}\right)\left(y-\left(-9\right)\right)$$

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

$$9y^{2}+75y-54=9\left(y-\frac{2}{3}\right)\left(y+9\right)$$

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

$$9y^{2}+75y-54=9\times \frac{3y-2}{3}\left(y+9\right)$$

Cancel out 3, the greatest common factor in 9 and 3.

$$9y^{2}+75y-54=3\left(3y-2\right)\left(y+9\right)$$

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

$$x ^ 2 +\frac{25}{3}x -6 = 0$$

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

$$r + s = -\frac{25}{3} rs = -6$$

Two numbers r and s sum up to -\frac{25}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{25}{3} = -\frac{25}{6}. 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>

$$r = -\frac{25}{6} - u s = -\frac{25}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -6

$$(-\frac{25}{6} - u) (-\frac{25}{6} + u) = -6$$

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

$$\frac{625}{36} - u^2 = -6$$

Simplify the expression by subtracting \frac{625}{36} on both sides

$$-u^2 = -6-\frac{625}{36} = -\frac{841}{36}$$

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

$$u^2 = \frac{841}{36} u = \pm\sqrt{\frac{841}{36}} = \pm \frac{29}{6}$$

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

$$r =-\frac{25}{6} - \frac{29}{6} = -9 s = -\frac{25}{6} + \frac{29}{6} = 0.667$$

Evaluate

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