Question 327727
As is, this one is not factorable in a nice way.

If it were {{{3x^2-8x+5}}}, then it would be (x-1)(3x-5)

But, as is, it has solutions {{{-(sqrt(31)-4)/3}}} and {{{(sqrt(31)+4)/3}}}.

Here is a trick to tell rather or not a quadratic is factorable or not before you spend time wrestling with it and it turns out it is not factorable.

Check the discriminant. It is {{{b^2-4ac}}}

You may recognize this as what is inside the parentheses in the quadratic formula. 

Anyway, a=3, b=-8, c=-5

{{{(-8)^2-4(3)(-5)=124}}}

If, when you check the discrimiant, the result is not a perfect square, then it is NOT factorable. 124 is not a perfect square, so it is not factorable.

The other case I mentioned, {{{3x^2-8x+5}}}

It has discriminant equal to 4. That is a perfect square. So it is factorable, as was shown. 

Like that?.