Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function. \(f(x) = 3x^3 + 39x^2 + 39x + 27\)
what does the rational root thrm tell us?
This is what my lesson says but I don't really get what it means. http://prntscr.com/4p0tjk
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
so, do you agree that 3 and 27 are the values we need to use?
yes
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
3 - 1,3 27 - 1, 3, 9, 27
so\[\pm\frac{1,3}{1,3,9,27}\] ignoring signs, there are at best 4(2) options for possible roots
in proper, 1/1 and 3/3 are equal so we wouldnt want to list them both, and just use the simplest forms
So is that all they want or is there more?
they want it formated into a proper list, but yeah, thats the brunt of it all
do you understand how to find the elements of the list?
1/1, 1/3, 1/9, 1/27, 3/1, 3/3, 3/9, 3/27?
yes, but its not 'proper' yet.
reduce and just list duplicate elements once
and include each term as postive and negative
ok and I need to use \[\pm \] in front of each one right?
i would, yes
\(\pm1, \pm1/3, \pm1/9, \pm1/27, \pm3,\)
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
Thank you so much for all of your help!
good luck
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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