anonymous
  • anonymous
According to the Rational Root Theorem, what are all the potential rational roots of f(x) = 5x^3 – 7x + 11?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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triciaal
  • triciaal
first you see the highest degree is 3 so there are 3 roots
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=f%28x%29+%3D+5x^3+%E2%80%93+7x+%2B+11
triciaal
  • triciaal
factor the polynomial if x-a is a factor when x = a there is no remainder

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anonymous
  • anonymous
Is 1 always a root?
triciaal
  • triciaal
|dw:1438798603916:dw|
zepdrix
  • zepdrix
List out the factors of your `leading coefficient`: Hmm so factors of positive 5 will be, 1, 5 and also -1, -5 Next, list out factors of your `constant term`: So factors of 11 will be, 1, 11 and also -1, -11 Rational root theorem tells us that IF the polynomial has a rational root, it will be a ratio of these factors. (The constant term's factor goes on top). Example: One possible rational root would be: \(\large\rm \frac{-11}{5}\)
zepdrix
  • zepdrix
Another possibility would be: \(\large\rm \frac{11}{1}\) which is simply \(\large\rm 11\). Does it uhhhhhhhh kinda make sense? +_+ They just want you to list them out.
triciaal
  • triciaal
@zepdrix why wouldn't it be 1/2 of that?
triciaal
  • triciaal
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anonymous
  • anonymous
could 5/11 be a possibility?
zepdrix
  • zepdrix
Hmmm here's an example:\[\large\rm x^2+4x+3\]Rational root theorem tells us to look at the factors. List of factors of the constant term: \(\large\rm 1,3\) and also \(\large\rm -1,-3\). List of factors of the leading conefficient: \(\large\rm 1\) and also \(\large\rm -1\). So our potention rational roots would be the different fractions with the constant factors on top. \[\large\rm \frac{-3}{1},\frac{-3}{-1},\frac{3}{1},\frac{3}{-1},\frac{1}{1},\frac{1}{-1},\frac{-1}{1},\frac{-1}{-1}\]I know I repeated some numbers in there, I just wanted to list out every possible root to maybe get the point across. So our rational root would have to come from that list of fractions.
zepdrix
  • zepdrix
You can easily factor that polynomial to find out that the roots are -3 and -1. Which come from that list, so I think it works out ok :)
zepdrix
  • zepdrix
5/11, no. The `constant factor` (which came from the number on the far right) has to go on top. Good question though :)
triciaal
  • triciaal
|dw:1438799505461:dw|
anonymous
  • anonymous
What about just plain 1 or 0? Or could that not work because it wasnt one of the constant factors
zepdrix
  • zepdrix
One of the factors of 11 is 1, that could be our numerator. One of the factors of 5 is 1, so that could be our denominator. So yes, a `potential` rational root could be 1/1.
triciaal
  • triciaal
oops I think I missed the key word rational can be expressed as ratio
zepdrix
  • zepdrix
To answer your earlier question: "Is 1 always a root?" It always has the `posibilitiy` of being a rational root. It's not always guaranteed to be an actual root though, as is the case with this problem. 1 is not a root of this polynomial that we have, but it was on our list of possible choices :)
zepdrix
  • zepdrix
So when you simplify all of your choices down (getting rid of repeats), you should have these for your options:\[\large\rm 1,11,-11,\frac{11}{5},-\frac{11}{5},\frac{1}{5},-\frac{1}{5}\] If you individually plugged these into the polynomial, you would find that none of them actually work out for you. So this poynomial has no rational roots. They don't want you to go that far though, they just want the list that I pasted above. I hate to spill the beans and just give you the answer, but this one is a little tricky :)
zepdrix
  • zepdrix
Hopefully I didn't miss any :d

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