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snowflakelove15 · Answer

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zepdrix · Answer

Hello Ms Snowflake c:

It turns out that none of these are `actual` roots.
But that's ok.
They just want us to apply our Rational Root Theorem to look for `possible` roots.

$\large\rm 5x^2-3x+3$
If we take a 5 out of everything,
$\large\rm 5\left(x^2-\frac35x+\frac35\right)$

In order for our second order polynomial to have rational roots,
they must be a factor of the constant term on the end.

zepdrix · Answer

$\large\rm 3\cdot\frac15=\frac35$

So we can certainly have the $\pm3$ but also the $\pm \frac15$ as possible roots.

But on the other hand,
$\large\rm 5\cdot\frac13\ne\frac{3}{5}$
It seems like these two numbers will not work, ya?

snowflakelove15 · Answer

Thank you very much! I totally get it now

zepdrix · Answer

There's another possible root, do you see why? c:

satellite73 · Answer

it says "possible" right, not "actual"