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hscishard

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There's a better way than substituting all the p/q values to find a rational root right? I use my calc for it, but it still takes forever!
So is this all about luck?

e.g. 2x^3 -1=0 potential roots are +\-1/2, +\-1 but a much longer one
 

Deep Blue

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There's a better way than substituting all the p/q values to find a rational root right? I use my calc for it, but it still takes forever!
So is this all about luck?

e.g. 2x^3 -1=0 potential roots are +\-1/2, +\-1 but a much longer one
Not that I know of. I'll have a look and see though.

EDIT: I just found this website and it seems to be helpful in cutting down some of the time taken.
http://www.purplemath.com/modules/solvpoly.htm
 
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jyu

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Not that I know of. I'll have a look and see though.

EDIT: I just found this website and it seems to be helpful in cutting down some of the time taken.
http://www.purplemath.com/modules/solvpoly.htm
This method of finding the rational roots works only when they exist. It works for most of the questions in textbooks because the questions were prepared for it to work, i.e. rational roots were used to construct the polynomials. Just think about this: does a polynomial always cross the x-axis at rational numbers?
 

Amogh

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This method of finding the rational roots works only when they exist. It works for most of the questions in textbooks because the questions were prepared for it to work, i.e. rational roots were used to construct the polynomials. Just think about this: does a polynomial always cross the x-axis at rational numbers?
Original question:
There's a better way than substituting all the p/q values to find a rational root right? I use my calc for it, but it still takes forever!
 

Drongoski

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This method of finding the rational roots works only when they exist. It works for most of the questions in textbooks because the questions were prepared for it to work, i.e. rational roots were used to construct the polynomials. Just think about this: does a polynomial always cross the x-axis at rational numbers?
Note this well and learn from it. Good work Mr jyu.
 
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hscishard

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This method of finding the rational roots works only when they exist. It works for most of the questions in textbooks because the questions were prepared for it to work, i.e. rational roots were used to construct the polynomials. Just think about this: does a polynomial always cross the x-axis at rational numbers?
ok yea, that was a bad example since I made it on the spot. Though I'm talking about when the question says there's at least 1 rational root...but you knew what I meant

I hate using the calculator but oh well
 

blackops23

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There's a better way than substituting all the p/q values to find a rational root right? I use my calc for it, but it still takes forever!
So is this all about luck?

e.g. 2x^3 -1=0 potential roots are +\-1/2, +\-1 but a much longer one
excuse me, quick question that I'm a tad curious about --> how do you know what the "potential roots are" in a non-monic polynomial? E.g. in a monic polynomial, the integer roots were factors of the constant. So what is the rule regarding rational roots (non-integer)?

Please explain. Thankyou.
 

seanieg89

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If p/q is a rational root of a polynomial with integer coefficients, then p must divide the constant term and q must divide the leading coefficient. For the monic case this reduces to the integer root condition you state above.
Also note that the 'p/q' here has to be in simplest form, i.e. p and q must have no common factors.
 

hscishard

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If p/q is a rational root of a polynomial with integer coefficients, then p must divide the constant term and q must divide the leading coefficient. For the monic case this reduces to the integer root condition you state above.
Also note that the 'p/q' here has to be in simplest form, i.e. p and q must have no common factors.
what he said
you can prove it easily by letting p/q being a root where p and q have no common factors.
 

blackops23

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If p/q is a rational root of a polynomial with integer coefficients, then p must divide the constant term and q must divide the leading coefficient. For the monic case this reduces to the integer root condition you state above.
Also note that the 'p/q' here has to be in simplest form, i.e. p and q must have no common factors.
aaahhhhhhhh..... thank you very much
 

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