# Composition of two polynomials

How's to make the composition of two polynomials? According to this page:

If $P = (x^3 + x)$, $Q = (x^2 + 1)$ then,

$P\circ Q = P\circ (x^2 + 1) = (x^2 + 1)^3 + (x^2 + 1) = x^6 + 3 x^4 + 4 x^2 + 2$

It seems that the $(x^3 + x)$ becomes the $x^3$, then we have $( \space \space \space )^3$ and now we just need to switch the inside of $P$ by the inside of $Q$ thus $(x^2 + 1)^3$.

I'm just not sure if my interpretation is correct. I'm also aware that I may not be using the right terms for describing this, but it's what I have now.

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$x^3+x$ doesn't become $x^3$, it stays as $x^3+x$ and then you have $(\ )^3+(\ )$. :) – Rahul Jul 23 '12 at 15:13
By definition $\: (P\circ Q)(x)\, =\, P(Q(x))\, =\, Q(x)^3 + Q(x)\ \$ – Bill Dubuque Jul 23 '12 at 15:39
I remember of studying Sum of two polynomials, difference of two polynomials, product of a constant and a polynomial and product of two polynomials. They kinda make sense for me, where is compostion of two polynomials useful? – Voyska Jul 23 '12 at 21:30
Ah, Another question: Is composition used only on polynomials with two constants? The given examples show me only operations on polynomials such as $(x^3 + x)$ but I've seen no references to polynomials with 3 constants such as $(x^3 + 4x^2 - x)$. – Voyska Jul 24 '12 at 4:58

Looks fine. Maybe it becomes even clearer, when you write it like: $$P\circ Q = (x^3 + x)\circ Q= (Q^3+Q)=(x^2 + 1)^3 + (x^2 + 1) = x^6 + 3 x^4 + 4 x^2 + 2$$

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You again, Draks? May our friendship last forever, While I'm studying mathematics, I'll ask a lot. Any doubt about our friendship? Axiom 1: There is a friendship between draks and GustavoB such that draks answers GustavoB's noob questions. – Voyska Jul 23 '12 at 21:14
@GustavoBandeira Thanks, but don't expect too much, I'm not even a mathematician. The time will come when you answer my questions. Good luck for your studies... – draks ... Jul 23 '12 at 21:28
Well. I can't even sum 2 integers - But I guess I can odd sum 2 integers. You're higher on the hierarchical tree - considering it exists. – Voyska Jul 24 '12 at 0:19