how would I factor
2x^3-23x^2+71x-30
and/or solve
500x^3+108=0
2x^3-23x^2+71x-30
and/or solve
500x^3+108=0
-
use the rational root theorem to find possible roots
factors of the constant divided by factors of the leading coefficient (plus or minus)
denominator = 1,2,3,5,6,10,15,30
numerator = 1,2
That is a lot to try, but after you find the first one that works, you will have a quadratic that you can solve the usual ways
(x-5)(x-6)(2x-1)
++++++
again use the rational root theorem.
I got (5x+3) as a factor and then a quadratic that is not factorable with real numbers.
factors of the constant divided by factors of the leading coefficient (plus or minus)
denominator = 1,2,3,5,6,10,15,30
numerator = 1,2
That is a lot to try, but after you find the first one that works, you will have a quadratic that you can solve the usual ways
(x-5)(x-6)(2x-1)
++++++
again use the rational root theorem.
I got (5x+3) as a factor and then a quadratic that is not factorable with real numbers.