Express in the form (x-a) 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 $f(a)$
 A: 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.
A: 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.  
