# Thread: Factoring Trinomial with fractions

1. ## Factoring Trinomial with fractions

Having issues with the following problem:

-1/3x^2 + 6x + 6 = y

I tried converting everything into fractions, but to no avail. Attempting to factor in order to find the vertex and axis of symmetry.

Any assistance that may be provided to lead me in the right direction would be greatly appreciated!

Thanks!

2. $\displaystyle y = -\frac{1}{3}x^2 + 6x + 6$

$\displaystyle = -\frac{1}{3}(x^2 - 18x - 18)$

$\displaystyle = -\frac{1}{3}\left[x^2 - 18x + (-9)^2 - (-9)^2 - 18\right]$

$\displaystyle = -\frac{1}{3}\left[(x-9)^2 - 81 - 18\right]$

$\displaystyle = -\frac{1}{3}\left[(x-9)^2 - 99\right]$

$\displaystyle = -\frac{1}{3}\left[(x-9)^2 - (\sqrt{99})^2\right]$

$\displaystyle = -\frac{1}{3}\left[(x - 9)^2 - (3\sqrt{11})^2\right]$

$\displaystyle = -\frac{1}{3}(x - 9 - 3\sqrt{11})(x - 9 + 3\sqrt{11})$.

3. Originally Posted by Hapa
Having issues with the following problem:

-1/3x^2 + 6x + 6 = y

I tried converting everything into fractions, but to no avail. Attempting to factor in order to find the vertex and axis of symmetry.

Any assistance that may be provided to lead me in the right direction would be greatly appreciated!

Thanks!
$-\frac{1}{3}(x^2 - 18x) = y - 6$

$-\frac{1}{3}(x^2 - 18x + 81) = y - 6 - 27$

$-\frac{1}{3}(x - 9)^2 = y - 33
$

$-\frac{1}{3}(x - 9)^2 + 33 = y$

can you finish?

4. my apologies for not stating the question correctly, what i meant to convey was:

What were the two binomials used to end up with:

-1/3x^2 + 6x+ 6

thanks again and sorry for the confusion

5. And my answer gives them...