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How to express $2x-1$ in the form $x-a$ ? Isn't $x-1/2$ wrong?

And how to express $2x+3$ in the form x-a?

this is for using it in the remainder theorem:

when $f(x)$ is divided by $x-a$, the remainder is $f(a)$

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I'm afraid that $2x-1=2(x-\frac12)$ is the best you can do. –  Andrea Mori Jul 23 '11 at 12:15
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$2x-1=2\left(x-\frac{1}{2}\right)$. What exactly you want to get? –  Dennis Gulko Jul 23 '11 at 12:15
    
remainder theorem $x-a$ –  gcse Jul 23 '11 at 12:19
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It seems that you are trying to have all the coefficient as integers, which is something that is not always possible. –  Asaf Karagila Jul 23 '11 at 15:37
    
@Asaf Probably the OP is working with polynomials over a field, e.g. the coefficient ring may be $\:\mathbb Q,\:\mathbb R\:$ or $\:\mathbb C\:.\:$ –  Bill Dubuque Jul 23 '11 at 17:47
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2 Answers

If you have $f(x)=2x-1$

then $f(x) = 2(x-a) + (2a-1)$

so the remainder on dividing $f(x)$ by $x-a$ is $2a-1$, which is $f(a)$.

Repeating this with $2x+3$ is similar.

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That's not the OP's question. Rather, the OP wants to know how to extend the remainder theorem to compute the remainder modulo non-monic linear polynomials. See my answer. –  Bill Dubuque Jul 23 '11 at 15:35
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@Bill: I rather suspect it is gcse's question. Look at the difficulty of another question. If you have 15 year old children in the UK then I am ready to be corrected. –  Henry Jul 23 '11 at 17:12
    
No, see his emphasized sentence above, and see also the duplicate question posted a few hours later - which leaves no doubt that the OP's goal is to find a way to apply the remainder theorem to non-monic linear divisors - as I said above. –  Bill Dubuque Jul 23 '11 at 17:40
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HINT $\ $ I presume that your question is how to use the Remainder Theorem to compute the remainder of polynomials modulo a linear polynomial $\rm\ b\:x-a\ $ that is non-monic, i.e. whose leading coefficient $\ne 1\:.\:$ In fact one can derive this non-monic case simply by scaling the result of the Remainder Theorem applied to the associated monic form $\rm\: x - a/b\:,\:$ essentially exploiting

$$\rm \dfrac{f(x)}{b\:x-a}\ =\ \dfrac{f(x)/b}{x-a/b}$$

Remainder Theorem $\rm\: \Rightarrow\ f(x)/b\ =\ q(x)\ (x-a/b) + c\quad\ $ for $\rm\quad\ \ c = f(a/b)/b$

Hence scaling by $\rm\:b\:$ yields$\rm\quad\ \ f(x)\: =\: q(x)\ (b\:x-a) + b\:c\quad$ for $\rm\quad b\:c = f(a/b)$

Hence the remainder of $\rm\:f(x)\:$ modulo a linear polynomial is $\rm\:f(r)\:,\:$ where $\rm\:r\:$ is the root of the linear polynomial. Said in congruence language $\rm\: mod\:\ x-r:\ \ x\equiv r\ \Rightarrow\ f(x)\equiv f(r)\:.\:$ This scaling trick works not only for polynomials over fields but over any coefficient ring where $\rm\:b\:$ is invertible.

The same idea also works for dividing by higher degree polynomials whose leading coefficients are units (invertible). For such polynomials the long-hand polynomial Division Algorithm always works, because the leading coefficient, being a unit, divides every other coefficient. If the leading coefficient is not a unit then division with remainder may not be possible. For example there is no $\rm\ q(x)\in \mathbb Z[x]\:,\ c\in \mathbb Z\ $ such that $\rm x\: =\: 2\:x\ q(x)+ c\:.\: $ If so, then evaluating at $\rm\: x = 0\ $ shows $\rm\ c = 0\:,\:$ then evaluating at $\rm\:x = 1\:$ shows $\rm\:1 = 2\:q(1)\:,\:$ so $2$ divides $1$ in $\rm\:\mathbb Z\:,\:$ a contradiction.

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