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Mathematics 15 Online
OpenStudy (sleepyjess):

Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function. \(f(x) = 3x^3 + 39x^2 + 39x + 27\)

OpenStudy (amistre64):

what does the rational root thrm tell us?

OpenStudy (sleepyjess):

This is what my lesson says but I don't really get what it means. http://prntscr.com/4p0tjk

OpenStudy (amistre64):

fair enough, in short, take the coefficent of the highest degree term, and the constant term, and use them to produce a pool of options

OpenStudy (amistre64):

so, do you agree that 3 and 27 are the values we need to use?

OpenStudy (sleepyjess):

yes

OpenStudy (amistre64):

then when we factor them and set them up as a rational number: factors of 3 ---------- factors of 27 the different combinations result in a pool of the only possible rational zeros that it can have

OpenStudy (sleepyjess):

3 - 1,3 27 - 1, 3, 9, 27

OpenStudy (amistre64):

so\[\pm\frac{1,3}{1,3,9,27}\] ignoring signs, there are at best 4(2) options for possible roots

OpenStudy (amistre64):

in proper, 1/1 and 3/3 are equal so we wouldnt want to list them both, and just use the simplest forms

OpenStudy (sleepyjess):

So is that all they want or is there more?

OpenStudy (amistre64):

they want it formated into a proper list, but yeah, thats the brunt of it all

OpenStudy (amistre64):

do you understand how to find the elements of the list?

OpenStudy (sleepyjess):

1/1, 1/3, 1/9, 1/27, 3/1, 3/3, 3/9, 3/27?

OpenStudy (amistre64):

yes, but its not 'proper' yet.

OpenStudy (amistre64):

reduce and just list duplicate elements once

OpenStudy (amistre64):

and include each term as postive and negative

OpenStudy (sleepyjess):

ok and I need to use \[\pm \] in front of each one right?

OpenStudy (amistre64):

i would, yes

OpenStudy (sleepyjess):

\(\pm1, \pm1/3, \pm1/9, \pm1/27, \pm3,\)

OpenStudy (amistre64):

let me verify 1/1, 1/3, 1/9, 1/27, 3/1, 3/3, 3/9, 3/27 1, 1/3, 1/9, 1/27, 3, 1, 1/3, 1/9 1, 1/3, 1/9, 1/27, 3 and to be proper we might need them sorted 1, 1/27, 1/9, 1/3, 3

OpenStudy (sleepyjess):

Thank you so much for all of your help!

OpenStudy (amistre64):

good luck

OpenStudy (triciaal):

I think there should be 6 @amistre64 do you think you missed one or do you think I have it wrong using the Descartes rule for signs f(x) + + + + no change in sign 3 roots none positive real f(-x) - + - + 3 changes in sign 3 real negative roots

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