What is the coefficient of $\sqrt{2}$ in $f(\sqrt{2})$ where $f(x) = 3 - x + 2x^2 - 5x^3 +4x^4?$

Question

What is the coefficient of $\sqrt2$ in $f(\sqrt2)$ where $f(x) = 3 - x + 2x^2 - 5x^3 +4x^4?$

Hi I'm studying polynomial rings. I don't get what the question is asking me to do. Should I just plug in $\sqrt2 $ in $f(x)$? Isn't the definition of coefficient the number in front of a variable? How can there be a coefficient of a number, i.e. $\sqrt2$?

Am I missing something..

PierreCarre · Answer 1 · 2020-10-28T13:51:04.717

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Well, when you compute $f(\sqrt{2})$ you'll obtain a result of the form $a + b \sqrt{2}$, where $a,b$ are integers. You are just being asked to say what $b$ is.

edited Oct 28 '20 at 13:51

answered Oct 28 '20 at 13:49

PierreCarre

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ah!!Thank you.. geez – jun Oct 28 '20 at 13:49

score 0 · Answer 2 · answered Oct 28 '20 at 18:25

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The square root doesn't survive even powers. As $-x-5x^3=-x(1+5x^2)$ the coefficient is $-(1+5\cdot2)$.

answered Oct 28 '20 at 18:25

Michael Hoppe

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i.e. if we split $,f,$ into even and odd parts $,f(x) = f(x^2) + x,h(x^2),$ then $,f(\sqrt 5) = f(5) + h(5)\sqrt 5\ \ $ – Bill Dubuque Oct 28 '20 at 21:59

What is the coefficient of $\sqrt{2}$ in $f(\sqrt{2})$ where $f(x) = 3 - x + 2x^2 - 5x^3 +4x^4?$

2 Answers2