Asked by kristie
how would you simplify y= (x^5-32)/(x-2)?
I know that it would equal (x^5-2^5)/(x-2), but what's the next step?
It wouldn't be just (x^4-2^4), would it?
kristie, in general I think you can prove (x-a)|(x^n - a^n); read (x-a) divides the expression on the right of '|'. You should be able to prove easily (x-a)*(x^(n-1) +ax^(n-2) +a^2 x^(n-3) + ... + a^(n-1)) = (x^n - a^n).
Here we have (x-2)|(x^5 - 2^5). Can you supply the details?
(Hopefully I recalled the correct identity; check this, it should be in your text.)
y= (x^5-32)/(x-2)?
y=(x^5-2^5)/(x-2)
You need to factor the numberator. which is likely not a easy task. However, if one of the factors is 2, you can make some headway. Checking to see if two is a factor..
y= (2^5-2^5)/(2-2)= ? 0/0 so it is not certain yet if y=0, and x=2 is a factor.
So divide the numberator by the denominator...
I did synthetic division to see if x=2 is a zero...you do the division the way you learned..
y= (x-2)(polynomial)/(x-2) and the x-2 factors divid out.
kristie you need to check the division I did, it is easy to make errors on those.
I am still confused about what I need to do...
Divide x-2 into the numerator, see if it goes evenly.
OK. I'll try that. Thank you so much for your help!
I know that it would equal (x^5-2^5)/(x-2), but what's the next step?
It wouldn't be just (x^4-2^4), would it?
kristie, in general I think you can prove (x-a)|(x^n - a^n); read (x-a) divides the expression on the right of '|'. You should be able to prove easily (x-a)*(x^(n-1) +ax^(n-2) +a^2 x^(n-3) + ... + a^(n-1)) = (x^n - a^n).
Here we have (x-2)|(x^5 - 2^5). Can you supply the details?
(Hopefully I recalled the correct identity; check this, it should be in your text.)
y= (x^5-32)/(x-2)?
y=(x^5-2^5)/(x-2)
You need to factor the numberator. which is likely not a easy task. However, if one of the factors is 2, you can make some headway. Checking to see if two is a factor..
y= (2^5-2^5)/(2-2)= ? 0/0 so it is not certain yet if y=0, and x=2 is a factor.
So divide the numberator by the denominator...
I did synthetic division to see if x=2 is a zero...you do the division the way you learned..
y= (x-2)(polynomial)/(x-2) and the x-2 factors divid out.
kristie you need to check the division I did, it is easy to make errors on those.
I am still confused about what I need to do...
Divide x-2 into the numerator, see if it goes evenly.
OK. I'll try that. Thank you so much for your help!
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