Solve 9y^2+6y-8

$$9 y ^ { 2 } + 6 y - 8$$

Factor

$$\left(3y-2\right)\left(3y+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9y^{2}+ay+by-8. To find a and b, set up a system to be solved.

$$a+b=6 ab=9\left(-8\right)=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 6.

$$a=-6 b=12$$

Rewrite 9y^{2}+6y-8 as \left(9y^{2}-6y\right)+\left(12y-8\right).

$$\left(9y^{2}-6y\right)+\left(12y-8\right)$$

Factor out 3y in the first and 4 in the second group.

$$3y\left(3y-2\right)+4\left(3y-2\right)$$

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

$$\left(3y-2\right)\left(3y+4\right)$$

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.

$$9y^{2}+6y-8=0$$

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.

$$y=\frac{-6±\sqrt{6^{2}-4\times 9\left(-8\right)}}{2\times 9}$$

Square 6.

$$y=\frac{-6±\sqrt{36-4\times 9\left(-8\right)}}{2\times 9}$$

Multiply -4 times 9.

$$y=\frac{-6±\sqrt{36-36\left(-8\right)}}{2\times 9}$$

Multiply -36 times -8.

$$y=\frac{-6±\sqrt{36+288}}{2\times 9}$$

Add 36 to 288.

$$y=\frac{-6±\sqrt{324}}{2\times 9}$$

Take the square root of 324.

$$y=\frac{-6±18}{2\times 9}$$

Multiply 2 times 9.

$$y=\frac{-6±18}{18}$$

Now solve the equation y=\frac{-6±18}{18} when ± is plus. Add -6 to 18.

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

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{-6±18}{18} when ± is minus. Subtract 18 from -6.

$$y=-\frac{24}{18}$$

Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.

$$y=-\frac{4}{3}$$

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 -\frac{4}{3} for x_{2}.

$$9y^{2}+6y-8=9\left(y-\frac{2}{3}\right)\left(y-\left(-\frac{4}{3}\right)\right)$$

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

$$9y^{2}+6y-8=9\left(y-\frac{2}{3}\right)\left(y+\frac{4}{3}\right)$$

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

$$9y^{2}+6y-8=9\times \frac{3y-2}{3}\left(y+\frac{4}{3}\right)$$

Add \frac{4}{3} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9y^{2}+6y-8=9\times \frac{3y-2}{3}\times \frac{3y+4}{3}$$

Multiply \frac{3y-2}{3} times \frac{3y+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9y^{2}+6y-8=9\times \frac{\left(3y-2\right)\left(3y+4\right)}{3\times 3}$$

Multiply 3 times 3.

$$9y^{2}+6y-8=9\times \frac{\left(3y-2\right)\left(3y+4\right)}{9}$$

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

$$9y^{2}+6y-8=\left(3y-2\right)\left(3y+4\right)$$

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

$$x ^ 2 +\frac{2}{3}x -\frac{8}{9} = 0$$

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

$$r + s = -\frac{2}{3} rs = -\frac{8}{9}$$

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

$$r = -\frac{1}{3} - u s = -\frac{1}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{8}{9}

$$(-\frac{1}{3} - u) (-\frac{1}{3} + u) = -\frac{8}{9}$$

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

$$\frac{1}{9} - u^2 = -\frac{8}{9}$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = -\frac{8}{9}-\frac{1}{9} = -1$$

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

$$r =-\frac{1}{3} - 1 = -1.333 s = -\frac{1}{3} + 1 = 0.667$$

Evaluate

$$\left(3y-2\right)\left(3y+4\right)$$