Polynomial division proof

In summary, Spivak's "Calculus," chapter 3 - problem 7 - a states that for any polynomial function f, and any number a, there is a polynomial function g, anad a number b, such that f(x) = (x - a)*g(x) + b for all x. The attempt at a solution is to prove this for a degree k+1, but fails because f(x) has degree k + 1 and cannot be combined with (x - a) to have a degree k polynomial.
  • #1
osnarf
209
0
Spivak's "Calculus," chapter 3 - problem 7 - a

Homework Statement


Prove that for any polynomial function f, and any number a, there is a polynomial function g, anad a number b, such that f(x) = (x - a)*g(x) + b for all x. (The idea is simply to divide (x - a) into f(x) by long division, until a constant remainder is left. For example, the calculation : see below - section 2 A formal proof is possible by induction on the degree of f.

Homework Equations


<here spivak divides x3 - 3x + 1 by x - 1 using long division>
<he arrives at x3 -3x +1 = (x - 1)(x2 + x -2) - 1>

The Attempt at a Solution



Proving a polynomial of degree 1 is easy enough for the induction proof.

For proving it for a degree k+1, spivak writes :
suppose the result is true for polynomials of degree <= k. If f has degree k + 1, then f has the form:

f(x) = ak + 1xk+1 + ... + a1 + a 0.


agreed so far...
then he writes, immediately following the previous quote:
Now the polynomial function h(x) = f(x) - ak+1(x-a) has degree <k, so we can...

The rest is simple, assuming the previous statement is correct. However, I don't understand how subtracting
ak+1(x-a)
from a function of degree k+1 is going to make it <k, unless k was 0.

Could anyone explain this to me please?
 
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  • #2
osnarf said:
Spivak's "Calculus," chapter 3 - problem 7 - a

Homework Statement


Prove that for any polynomial function f, and any number a, there is a polynomial function g, anad a number b, such that f(x) = (x - a)*g(x) + b for all x. (The idea is simply to divide (x - a) into f(x) by long division, until a constant remainder is left. For example, the calculation : see below - section 2


A formal proof is possible by induction on the degree of f.


Homework Equations


<here spivak divides x3 - 3x + 1 by x - 1 using long division>
<he arrives at x3 -3x +1 = (x - 1)(x2 + x -2) - 1>

The Attempt at a Solution



Proving a polynomial of degree 1 is easy enough for the induction proof.

For proving it for a degree k+1, spivak writes :
suppose the result is true for polynomials of degree <= k. If f has degree k + 1, then f has the form:

f(x) = ak + 1xk+1 + ... + a1 + a 0.


agreed so far...
then he writes, immediately following the previous quote:
Now the polynomial function h(x) = f(x) - ak+1(x-a) has degree <k, so we can...
I think you are missing an exponent. The only way this makes sense is by subtracting ak+1xk+1. Then, h(x) is a polynomial of degree <= k.
Edit: changed the above from ak+1(x - a)k+1.
osnarf said:
The rest is simple, assuming the previous statement is correct. However, I don't understand how subtracting
ak+1(x-a)
from a function of degree k+1 is going to make it <k, unless k was 0.

Could anyone explain this to me please?
 
Last edited:
  • #3
okay i guess I'm not going crazy. that's the 3rd typo I've found in this book tonight
 
  • #4
wait wait... then the rest of the proof doesn't make sense. next it says:

...so we can write:

f(x) - ak+1(x - a) = (x - a)g(x) + b

or

f(x) = (x - a)[g(x) + ak+1] + b


which is the required form.


so...? Can't combine the (x - a)'s if one is to the k+1th power
 
  • #5
I'm not following that step either. f(x) is a degree k + 1 polynomial, so subtracting ak+1(x - a) doesn't get you to a degree k polynomial.

Can you include the full text of what you're looking at, consisting of the induction hypothesis and what follows?
 

What is polynomial division proof?

Polynomial division proof is a mathematical process used to prove that one polynomial is divisible by another. It involves dividing the polynomial into smaller parts and then showing that each part is a factor of the original polynomial.

How is polynomial division proof different from regular polynomial division?

Regular polynomial division is used to find the quotient and remainder when dividing one polynomial by another. Polynomial division proof goes a step further and shows that the remainder is equal to zero, proving that the polynomial is divisible.

What are the steps for conducting a polynomial division proof?

The steps for conducting a polynomial division proof are:1. Write the dividend (the polynomial being divided) and the divisor (the polynomial dividing the dividend) in descending order of powers.2. Divide the first term of the dividend by the first term of the divisor.3. Multiply the result by the entire divisor.4. Subtract this result from the first term of the dividend.5. Repeat the process with the new dividend until all terms have been divided.6. If the final result is equal to zero, then the polynomial is divisible and the proof is complete.

Why is polynomial division proof important?

Polynomial division proof is important because it allows us to determine whether one polynomial is a factor of another. This can be useful in solving equations, factoring polynomials, and understanding the relationships between different polynomials.

What are some common mistakes made when conducting polynomial division proof?

Some common mistakes when conducting polynomial division proof include:1. Forgetting to write the terms of the dividend and divisor in descending order.2. Making a mistake in the division step.3. Forgetting to multiply the result by the entire divisor.4. Making a mistake in the subtraction step.5. Not completing all the necessary steps.It is important to carefully follow each step and double check for any errors to ensure an accurate proof.

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