Polynomial function - different degrees don't understand

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Homework Help Overview

The discussion revolves around understanding polynomial functions, specifically the implications of subtracting a term from a polynomial of degree k+1 to analyze the resulting degree of the new polynomial function h(x).

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the form of polynomial functions and the effects of subtracting a specific term on the degree of the resulting polynomial. Questions arise regarding the cancellation of terms and the definition of polynomial degrees.

Discussion Status

Some participants provide insights into the polynomial structure and the degree concept, while others express confusion about the implications of the subtraction. There is a recognition of differing interpretations regarding the degree of h(x), with some suggesting corrections to earlier statements.

Contextual Notes

Participants are navigating through definitions and properties of polynomial degrees, with some expressing uncertainty about the notation and terminology used in the original problem statement.

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

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

Now obviously at this point the [tex]a_{k+1}x[/tex] will cancel with that from the polynomial f(x) and [tex]a\cdot a_{k+1}[/tex] is just some constant we can add to that of a0 from f(x). See now that the degree of h(x) will be [tex]\leq k[/tex] 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

[tex]2x^3-3x+1=0[/tex] has a coefficient of 0 for its 2nd degree.
 
Mentallic said:
Now obviously at this point the [tex]a_{k+1}x[/tex] 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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