# Question #f0544

Sep 15, 2017

$x = - \frac{1}{3}$

$x = \frac{3}{8}$

#### Explanation:

You want to get all terms on the same side, with the term with the highest degree being positive $\left(24 {x}^{2}\right)$.

So the new equation looks like

$24 {x}^{2} - x - 3 = 0$

Since you cannot factor anything, you have to use the AC method (multiplying the A times the C). You then have to figure out what two numbers multiply to your product of your AC, and add up to your B $\left(- x\right)$.

Those numbers are $8$ and $9$, more specifically, positive $8$ and $- 9$. You then replace your B with your to multiples of AC, so your new problem is:

$24 {x}^{2} - 9 x + 8 x - 3 = 0$

Now put them in their respective pairs

$\left(24 {x}^{2} - 9 x\right) + \left(8 x - 3\right) = 0$

Now factor out the GCF (Greatest Common Factor) possible, your two sets of parenthesis should be exactly the same! Your new equation is

$3 x \left(8 x - 3\right) + \left(8 x - 3\right) = 0$

Since there is an imaginary 1 in front of the second set of parenthesis, don't forget about it! Now put the GCF's in one set of parentheses and the current set of parentheses in another.

$\left(3 x + 1\right) \left(8 x - 3\right) = 0$

Set each set of parentheses equal to zero, and solve for $x$, those are your zeros/solutions/x-intercepts/etc.

P.S. - To check if you're right, you ca FOIL out the binomials and see if you get the original problem, if so, congratulations, if not, maybe try again :D!