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I have the equation $$x^3-8x^2+30x-20=0$$ let's call the roots $a,b,c$

It's easy to find the equation with roots $(a+2)(b+2)(c+2)$ these are the steps in my books

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so the equation is $$ aX^3+ bX^2 + cX + D = 0 $$ $$ x^3 -14x^2 +74x -120 = 0$$

my problem is, I can't understand why divide by $-2$ not $2$ how do I do this division with long division? like what am I dividing by if the symbols were here?!

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3 Answers 3

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You are finding $P(x-2)$, given $P(x)$. If $a$ is a root of $P(x)$, clearly $a+2$ is a root of $P(x-2)$.

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  • $\begingroup$ well, I'm disappointed in my self but it's not 'clearly' for me, I'll be thankful if you provide a more detailed answer $\endgroup$
    – Ahmed I. Elsayed
    May 5, 2019 at 3:04
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    $\begingroup$ @AhmedI.Elsayed Just replace $x$ in $P(x-2)$ with $a+2$, you will get $P(a)$, which is zero if $a$ is a root of $P(x)$. Try it and let me know if there is some confusion. $\endgroup$
    – Macavity
    May 5, 2019 at 3:06
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    $\begingroup$ Well, if $P(x)=0$, for $P((x+2))=0$, then it must be $P((x+2)-2)$, for $(x+2)-2=x$. $\endgroup$
    – DUO Labs
    May 5, 2019 at 3:07
  • $\begingroup$ Thanks, this will be appreciated $\endgroup$
    – Ahmed I. Elsayed
    May 5, 2019 at 3:23
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    $\begingroup$ Thank you for discussion and the reference. $\endgroup$
    – farruhota
    May 5, 2019 at 7:06
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Your book is using synthetic division. As you are dividing by $(x+2)$, the divisor must be $-2$. Generalized, if you are dividing polynomial $f(x)$ by $x+a$, for synthetic division, you use $-a$, which is exactly what your book did.

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  • $\begingroup$ surely i understand that, I got it anyway thanks $\endgroup$
    – Ahmed I. Elsayed
    May 5, 2019 at 3:22
  • $\begingroup$ As Quote Dave mentions, you are actually evaluation $P(-2)$ say to get the constant coefficient, which is the same as finding the remainder when you divide by $x+2$, which is the first synthetic division step you do in the diagram. Similarly from the quotient, you extract the next remainder, which is the coefficient of the linear term and so on. +1 $\endgroup$
    – Macavity
    May 5, 2019 at 3:25
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Alternatively: for the roots $a,b,c$ of $x^3-8x^2+30x-20=0$, by Vieta's we have: $$a+b+c=8\\ ab+bc+ca=30\\ abc=20$$ Now we want the roots $a+2,b+2,c+2$ of $X^3+kX^2+lX+m=0$. Hence: $$a+2+b+2+c+2=a+b+c+6=14=-k\\ (a+2)(b+2)+(b+2)(c+2)+(c+2)(a+2)=\\ (ab+bc+ca)+4(a+b+c)+12=74=l\\ (a+2)(b+2)(c+2)=\\ abc+2(ab+bc+ca)+4(a+b+c)+8=120=-m \Rightarrow \\ x^3-14x^2+74x-120=0.$$

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  • $\begingroup$ Vieta approach is more generic and works for symmetric polynomial transformations +1 $\endgroup$
    – Macavity
    May 5, 2019 at 7:02
  • $\begingroup$ Thank you, that is what I observed. $\endgroup$
    – farruhota
    May 5, 2019 at 7:04
  • $\begingroup$ Awesome approach thanks $\endgroup$
    – Ahmed I. Elsayed
    May 6, 2019 at 1:39
  • $\begingroup$ but tbh, it's not practical it will take a year to do this for a 4th or 5th degree equation $\endgroup$
    – Ahmed I. Elsayed
    Jun 6, 2019 at 0:50
  • $\begingroup$ Yes, you are right. Also as conmented above it will work for symmetric polynomial, i.e. it will not work when roots are $2,3,4$, etc. Anyway, it is a valid and guaranteed method. $\endgroup$
    – farruhota
    Jun 6, 2019 at 5:45

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