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 June 26th, 2012, 12:37 PM #1 Member   Joined: Apr 2012 Posts: 92 Thanks: 0 Polynomial help I would like to clarify one thing, in questions like, Find all third degree real polynomials with zeros of -1/2 and 1 -3i there's always this constant multiplied to the answer, such that the polynomial f(x) = a(answer). What and why is there a need for a constant to be multiplied? In a general polynomial like 3x^2+x+5, I don't see the need for existence of a constant, unless the constant is 1. Thank you!
 June 26th, 2012, 12:50 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Polynomial help Consider that the roots are not changed when multiplying the polynomial by a constant but the polynomial is changed. For instance: $x-1\ne2x-2$ Yet, the two binomials have the same root.
June 26th, 2012, 01:07 PM   #3
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Re: Polynomial help

Hello, Tutu!

Quote:
 I would like to clarify one thing, in questions like, Find all third degree real polynomials with zeros of -1/2 and 1 -3i. There's always this constant multiplied to the answer, such that the polynomial f(x) = a(answer). What and why is there a need for a constant to be multiplied? In a general polynomial like 3x^2+x+5, I don't see the need for existence of a constant, unless the constant is 1. Thank you!

$\text{The quadratic polynomial }\,P(x) \:=\:x^2\,-\,1\,\text{ has zeros of }+1\text{ and }-1.$

$\text{But is it the only one?}$

$\text{No . . . all these polynomials have the same zeros.}$

[color=beige]. . [/color]$\begin{array}{c}2x^2\,-\,2 \\ \\ \\ 37x^2\,-\,37 \\ \\ \\ 483x^2\,-\,483 \\ \\ \\ \vdots \end{array}$

$\text{Obviously, they are all multiple of }P(x).$

$\text{If they ask for }all\text{ quadratic polynomials with zeros }\pm 1
\;\;\text{we must answer: }\,c(x^2\,-\,1)\,\text{ for any real number }c\,\ne\,0.$

 June 26th, 2012, 01:10 PM #4 Member   Joined: Apr 2012 Posts: 92 Thanks: 0 Re: Polynomial help I see now, thank you thank you thank you!

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