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anonymous
 one year ago
According to the Rational Root Theorem, what are all the potential rational roots of f(x) = 5x^3 – 7x + 11?
anonymous
 one year ago
According to the Rational Root Theorem, what are all the potential rational roots of f(x) = 5x^3 – 7x + 11?

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triciaal
 one year ago
Best ResponseYou've already chosen the best response.1first you see the highest degree is 3 so there are 3 roots

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=f%28x%29+%3D+5x^3+%E2%80%93+7x+%2B+11

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1factor the polynomial if xa is a factor when x = a there is no remainder

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438798603916:dw

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1List 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
 one year ago
Best ResponseYou've already chosen the best response.1Another 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
 one year ago
Best ResponseYou've already chosen the best response.1@zepdrix why wouldn't it be 1/2 of that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0could 5/11 be a possibility?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmmm 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
 one year ago
Best ResponseYou've already chosen the best response.1You 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
 one year ago
Best ResponseYou've already chosen the best response.15/11, no. The `constant factor` (which came from the number on the far right) has to go on top. Good question though :)

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438799505461:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What about just plain 1 or 0? Or could that not work because it wasnt one of the constant factors

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1One 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
 one year ago
Best ResponseYou've already chosen the best response.1oops I think I missed the key word rational can be expressed as ratio

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1To 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
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.1Hopefully I didn't miss any :d
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