Solve 8g^2+6g-9

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

Factor

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

Solution Steps

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

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

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

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

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

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

Factor out common term 4g-3 by using distributive property.

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

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.

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

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.

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

Square 6.

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

Multiply -4 times 8.

$$g=\frac{-6±\sqrt{36-32\left(-9\right)}}{2\times 8}$$

Multiply -32 times -9.

$$g=\frac{-6±\sqrt{36+288}}{2\times 8}$$

Add 36 to 288.

$$g=\frac{-6±\sqrt{324}}{2\times 8}$$

Take the square root of 324.

$$g=\frac{-6±18}{2\times 8}$$

Multiply 2 times 8.

$$g=\frac{-6±18}{16}$$

Now solve the equation g=\frac{-6±18}{16} when ± is plus. Add -6 to 18.

$$g=\frac{12}{16}$$

Reduce the fraction \frac{12}{16} to lowest terms by extracting and canceling out 4.

$$g=\frac{3}{4}$$

Now solve the equation g=\frac{-6±18}{16} when ± is minus. Subtract 18 from -6.

$$g=-\frac{24}{16}$$

Reduce the fraction \frac{-24}{16} to lowest terms by extracting and canceling out 8.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{4} for x_{1} and -\frac{3}{2} for x_{2}.

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

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

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

Subtract \frac{3}{4} from g by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

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

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

Multiply 4 times 2.

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

Cancel out 8, the greatest common factor in 8 and 8.

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

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 8

$$x ^ 2 +\frac{3}{4}x -\frac{9}{8} = 0$$

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

$$r + s = -\frac{3}{4} rs = -\frac{9}{8}$$

Two numbers r and s sum up to -\frac{3}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{4} = -\frac{3}{8}. 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>

$$r = -\frac{3}{8} - u s = -\frac{3}{8} + u$$

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

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

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

$$\frac{9}{64} - u^2 = -\frac{9}{8}$$

Simplify the expression by subtracting \frac{9}{64} on both sides

$$-u^2 = -\frac{9}{8}-\frac{9}{64} = -\frac{81}{64}$$

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

$$u^2 = \frac{81}{64} u = \pm\sqrt{\frac{81}{64}} = \pm \frac{9}{8}$$

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

$$r =-\frac{3}{8} - \frac{9}{8} = -1.500 s = -\frac{3}{8} + \frac{9}{8} = 0.750$$

Evaluate

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