At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
in f(x), how many sign changes are there?
I don't know?
notice that going from -8x^4 to +25x^3, there's a sign change from negative to positive
do you see this?
ok there's another from +25x^3 to -8x^2
what's another one?
+x to -19?
there's one more
from -8x^2 to +x
so there are 4 sign changes total in f(x)
this means that there are at most 4 positive real roots
-19 to +x?
no that's the same sign change (just in reverse)
4 sign changes=4 positive real roots
at most 4 (there could be 0, 1, 2, 3, or 4 positive real roots)
4 is the maximum
now we must find f(-x)
I'm still really confuzzled...
All of it
the rule is if f(x) has n sign changes, then there are AT MOST n positive real roots
if n = 4, then if f(x) has 4 sign changes, then there are AT MOST 4 positive real roots
making more sense?
I don't know how to formulate an answer from you telling me this...
you don't need to find the actual roots you just need to be able to count the possible number and type
more specifically, how many positive real roots you could have (in this case, at most 4) you don't need to find the 4 or find out how many actual positive roots there are
But I need a definite answer, right?
I can't just say "maybe 4"?
well that's part of the answer, we still have to find f(-x)
Okay, and how would we go about that ? :)
start with f(x) then replace each x with -x and simplify
Could you give me an example?
start with this f(x) = –3x^5 – 8x^4 +25x^3 – 8x^2 +x – 19 then replace each 'x' with '-x' to get f(-x) = –3(-x)^5 – 8(-x)^4 +25(-x)^3 – 8(-x)^2 +(-x) – 19 then simplify to get ???
does that help?
Yes, it does :)
ok what do you get
when you simplify
f(-x) = –3(-x)^5 – 8(-x)^4 +25(-x)^3 – 8(-x)^2 +(-x) – 19 would simplify to f(-x) = 3x^5 – 8x^4 - 25x^3 – 8x^2 - x – 19
now count the sign changes in f(-x) = 3x^5 – 8x^4 - 25x^3 – 8x^2 - x – 19 and tell me how many you got
No sign changes?
3x^5 is the same as +3x^5
so there's only one sign change from +3x^5 to -8x^4
so this means that there is at most 1 negative real root
Oh, I was just looking at it wrong.. Sorry :(
So far we found that there are at most 4 positive real roots and at most 1 negative real root
so one possibility is that there are 4+1 =5 real roots (4 positive, 1 negative) BUT this is not the only possibility
we could have 3 positive real roots and 1 negative real root to have 3+1 = 4 real roots total but the problem with this scenario is that you would only have 5-1 = 1 complex root...when complex roots should come in pairs
so that scenario of 3 positive real roots and 1 negative real root just isn't possible
Okay So don't worry about the last scenario?
anyways, there are a ton of possibilities the good news is that you can sum them all up by saying there are at most 4 positive real roots (you could have 0, 1, 2, 3, or 4 positive real roots) there is at most 1 negative real root (you could have 0 or 1 negative real roots) there are at most 5 complex roots (if there are no real roots at all)
in short there are at most 4 positive real roots there is at most 1 negative real root there are at most 5 complex roots
I got it! :D
we can stop there because we don't need to find the actual roots or the counts of the types or roots...just the max of all of the possible types of roots