Understanding q_n(x) Expansions

[Jbreezy]

Homework Statement

Hi so I'm not understanding my reading of this text.

Homework Equations

Below is what I don't understand.

The Attempt at a Solution

$q_n(x) = (x-a_1)(x-a_2)(x-a_3)...(x-a_n)$
Now if we expanded this factorization by multiplying it out it should be clear the coefficient of $x^n$ should be one because it could only be derived by choosing x from each of the n bracket terms when composing the product.

So what I don't understand what this is telling me. Expand the factorization.
$q_n(x) = (x-a_1)(x-a_2)(x-a_3)...(x-a_n)$

For the first three terms I got

$(-x^3a_3 + x^2a_2a_3 + x^2a_1a_3 - a_1a_2a_3x ) ...(x-a_n)$
Not sure how to include the nth term. How to write it.

Would it be

$(a_3...a_nx^n - a_2a_3..a_nx^n -a_1a_3..a_nx^n + a_1a_2a_3..a_nx^n)$

seems like crap. I don't get what this is telling me!

 

[kevinferreira]

It is telling you that the term with x^n after doing the multiplication comes with a factor of 1 because it is the result of multiplying the n x's and all of them have a factor of 1! And 1^n=1.

Try with n=2 and then n=3.

 

[Mandelbroth]

Homework Statement

Hi so I'm not understanding my reading of this text.

Homework Equations

Below is what I don't understand.

The Attempt at a Solution

$q_n(x) = (x-a_1)(x-a_2)(x-a_3)...(x-a_n)$
Now if we expanded this factorization by multiplying it out it should be clear the coefficient of $x^n$ should be one because it could only be derived by choosing x from each of the n bracket terms when composing the product.

So what I don't understand what this is telling me. Expand the factorization.
$q_n(x) = (x-a_1)(x-a_2)(x-a_3)...(x-a_n)$

For the first three terms I got

$(-x^3a_3 + x^2a_2a_3 + x^2a_1a_3 - a_1a_2a_3x ) ...(x-a_n)$



Not sure how to include the nth term. How to write it.

Would it be

$(a_3...a_nx^n - a_2a_3..a_nx^n -a_1a_3..a_nx^n + a_1a_2a_3..a_nx^n)$

seems like crap. I don't get what this is telling me!

I get $(x-a_1)(x-a_2)(x-a_3)\cdots(x-a_n)\\=(x^2-(a_1+a_2)x+a_1a_2)(x-a_3)\cdots(x-a_n)\\=(x^3-(a_1+a_2+a_3)x^2+(a_1a_2+a_2a_3+a_1a_3)x-a_1a_2a_3)\cdots(x-a_n).$

 

[Jbreezy]

I get $(x-a_1)(x-a_2)(x-a_3)\cdots(x-a_n)\\=(x^2-(a_1+a_2)x+a_1a_2)(x-a_3)\cdots(x-a_n)\\=(x^3-(a_1+a_2+a_3)x^2+(a_1a_2+a_2a_3+a_1a_3)x-a_1a_2a_3)\cdots(x-a_n).$

How did you x^3 with no coefficient in front of it. I think you rearranged inbetween the steps.
How did you get this.

$(x^3-(a_1+a_2+a_3)x^2+(a_1a_2+a_2a_3+a_1a_3)x-a_1a_2a_3)\cdots(x-a_n).$

More steps. I see how you got the middle two parts but how do you have the last term before the nth as just coefficients?

 

[Jbreezy]

DOnt; answer this question thanks cya