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Consider this equation: $9(x-0.4)^4+2$. I just want to confirm that this is a 4th degree polynomial.

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closed as off-topic by Adam Hughes, Claude Leibovici, Watson, S.C.B., colormegone Apr 4 '16 at 7:18

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  • $\begingroup$ What leads you to think that it is or that it isn't? $\endgroup$ – Eric Towers Apr 4 '16 at 5:20
  • $\begingroup$ I am about 98% sure that it would be a 4th degree polynomial because it is raised to the 4th power. I just didn't know if the parentheses would change anything. $\endgroup$ – user328399 Apr 4 '16 at 5:23
  • $\begingroup$ The binomial theorem guarantees that the parentheses do not change anything. The degree of $(a+x)^n$ is $n$. $\endgroup$ – Wouter Apr 4 '16 at 5:32
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We can just do it by brute force: $$ 9\left(x-\frac{2}{5}\right)^4 + 2 = 9x^4 - \frac{72}{5}x^3 + \frac{216}{25}x^2 - \frac{288}{125}x + \frac{1394}{625} $$ or, you can notice that raising a polynomial of order $m$ to exponent $n$ yields a polynomial of degree $mn$. It can be shown rigorously using the Binomial Theorem.

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Yes , degree of a polynomial is the highest degree of its terms. In this case, it is indeed 4

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Yes, it is. In general, a function of the form

$f(x) = a(x-b)^n + c$ where $n\geq1$ and $a \neq 0$ is a polynomial of degree $n$. This can be verified with the binomial theorem.

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  • $\begingroup$ Thank you. And coefficients would be -0.4 and 2, right? $\endgroup$ – user328399 Apr 4 '16 at 5:24
  • $\begingroup$ @user328399 No. The leading coefficient would be $9$ $\endgroup$ – MathematicsStudent1122 Apr 4 '16 at 5:26
  • $\begingroup$ I meant to say constant, sorry mate. But, -0.4 isn't a constant? $\endgroup$ – user328399 Apr 4 '16 at 5:27
  • $\begingroup$ @user328399 No, the constant would be something else. Try expanding it for youself! $\endgroup$ – MathematicsStudent1122 Apr 4 '16 at 5:29
  • $\begingroup$ @MathematicsStudent1122 I think OP is asking if in their case $a=9$, $b=0.4$, $c=2$, in which case they're correct $\endgroup$ – leibnewtz Apr 4 '16 at 5:32

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