Can Calculus Proofs Clarify Polynomial Degrees and Factoring Challenges?

[Psybroh]

Hi, I've been trying a couple of proofs that my calc teacher gave me, but I'm not sure if I have the right approach or not.

1) Prove that the degree of the depressed polynomial is exactly one less than the degree of the original polynomial.

- For this proof, all I can come up is the face that since a "x" has been removed from P(x), the depressed polynomial Q(x) has 1 less "x" in each of its terms, and therefore has one less degree. is this correct? I cannot seem to express this in terms of variables and numbers.

2) Factor P(x) = x^3 + x^2 - 16x + 20 into the product of a constant and 3 linear factors.

- Can I factor out a 4 and use that as a constant?

3) Show that x=a is a root of x^3 - ax^2 + ax - mx^2 - a^2 + amx = 0, without using synthetic division.

- Do I just plug in x=a into the polynomial and show that the entire thing does come out to zero, or...?

4) Given that x=a is a root of x^3 - ax^2 - ax - mx - mx^2 + a^2 + am + amx = 0, use synthetic division to factor that eight-term polynomial into the product of two factors: one in linear x, one quadratic in x.

- I have NO idea how to even start this... any help would be great! lol

Well, these are it. I have some ideas to solve them, but I'm not sure if they can really be considered proofs. Any help and suggestions would be great! Thanks! :)

 

[LCKurtz]

Hi, I've been trying a couple of proofs that my calc teacher gave me, but I'm not sure if I have the right approach or not.

1) Prove that the degree of the depressed polynomial is exactly one less than the degree of the original polynomial.

- For this proof, all I can come up is the face that since a "x" has been removed from P(x), the depressed polynomial Q(x) has 1 less "x" in each of its terms, and therefore has one less degree. is this correct? I cannot seem to express this in terms of variables and numbers.

2) Factor P(x) = x^3 + x^2 - 16x + 20 into the product of a constant and 3 linear factors.

- Can I factor out a 4 and use that as a constant?

Of course not. 4 is not a common factor. Use the rational roots theorem and synthetic division.

3) Show that x=a is a root of x^3 - ax^2 + ax - mx^2 - a^2 + amx = 0, without using synthetic division.

- Do I just plug in x=a into the polynomial and show that the entire thing does come out to zero, or...?

Did you try that? Did it work?

4) Given that x=a is a root of x^3 - ax^2 - ax - mx - mx^2 + a^2 + am + amx = 0, use synthetic division to factor that eight-term polynomial into the product of two factors: one in linear x, one quadratic in x.

- I have NO idea how to even start this... any help would be great! lol

Collect terms on powers of x and use synthetic division by (x-a).

 

[Simon Bridge]

(1) ... what is the definition of a depressed polynomial?
Most proofs start out with a statement of definition.

(2) ... what happens when you try?
(As LCKurtz points out, 4 is not a common factor. Had you tried, you'd have got your answer.)
Sometimes the only way to identify the right path is to follow it for a while.
You would be better to try to factorize the polynomial first.

(3) ... what is the definition of "root of a polynomial"?
The comment here is a combination of those for (1) and (2) ... try it and see.

(4) if "a" is a root of the polynomial, then "(x-a)" is a factor.
How would you normally go about factorizing a polynomial?

This is giving me deja-vu ... anyway, LCKurtz has given you some good clues if you don't know some of the theory. It is all stuff you can look up. Let us know how you got on.