Polynomial function - different degrees don't understand

AI Thread Summary
The discussion centers on understanding polynomial functions and their degrees, specifically how the degree of a polynomial function h(x) can be less than or equal to k when derived from a polynomial f(x) of degree k+1. The confusion arises from the subtraction of the term ak+1(x-a), which eliminates the k+1 degree term, thus lowering the overall degree of h(x). Participants clarify that when the k+1 term cancels out, the highest remaining degree in h(x) is k or lower. The conversation highlights the importance of correctly defining polynomial degrees and their implications on function behavior. Overall, the key takeaway is that the degree of a polynomial is determined by its highest exponent, and specific operations can change that degree.
zeion
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Homework Statement



Hello.
I don't understand this:

Let f(x) be a polynomial function of degree k+1, then f(x) has the form
ak+1xk+1 + ... + a1x + a0

Now the polynomial function has degree h(x) = f(x) - ak+1(x-a) has degree <= k
How?



Homework Equations





The Attempt at a Solution



I can understand if (x-a) = xk+1 then the k+1 th term will be gone. But what does (x-a) do?
 
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Just think of (x-a) as (x-2) or (x-10), a being any number, just as ak and a2 etc. are any constants.

So h(x)=f(x)-a_{k+1}(x-a), now we can just expand out and simplify:

h(x)=f(x)-a_{k+1}x+a\cdot a_{k+1}

Now obviously at this point the a_{k+1}x will cancel with that from the polynomial f(x) and a\cdot a_{k+1} is just some constant we can add to that of a0 from f(x). See now that the degree of h(x) will be \leq k simply because we have eliminated the (k+1)th degree and the kth degree could have a coefficient of zero, just like the cubic polynomial

2x^3-3x+1=0 has a coefficient of 0 for its 2nd degree.
 
Mentallic said:
Now obviously at this point the a_{k+1}x will cancel with that from the polynomial f(x)

So x has to equal xk+1 ?
 
zeion said:

Homework Statement



Hello.
I don't understand this:

Let f(x) be a polynomial function of degree k+1, then f(x) has the form
ak+1xk+1 + ... + a1x + a0

Now the polynomial function has degree h(x) = f(x) - ak+1(x-a) has degree <= k
How?
As stated above, the degree of h(x) is still k + 1. I think you have a typo or there is is typo in what you're getting this from.

You have f(x) = ak+1xk+1 + ... + a1x + a0, and h(x) = f(x) - ak+1(x - a). This means that h(x) = ak+1xk+1 + ... + a1x - ak+1x + a*ak+1+ a0.

I believe that h(x) should be defined this way: h(x) = f(x) - ak+1(x-a)k + 1[/color]. If so, then the two xk + 1 terms cancel, and you're left with a polynomial whose highest-degree term is xk, making its degree <= k.

zeion said:

Homework Equations





The Attempt at a Solution



I can understand if (x-a) = xk+1 then the k+1 th term will be gone. But what does (x-a) do?
 
I'd like to make things simple for you the degree of a polynomial is it's highest exponent. Generally Trinomial + is considered a higher degree.

x^2 = Binomial
x^3 = Trinomial

Also another hint about degrees
1. They tell you the overall shape of the graph
2. They tell you how many x values there is
 
SpeedOfDark said:
I'd like to make things simple for you

Thanks I never knew things could be so simple.
 
Oh man I should really avoid posting when I'm sleepy... zeion as you defined h(x) in your OP, it isn't true that it will be of degree less than or equal to k. Mark44 fixed it up from there.
Sorry about that.
 
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