Retina 0.8.2, 56 bytes

(?=( \S+)+) x^$#1 \b0x.\d+  \b1x x x.1  x  0 - -   + 

Try it online! Link includes test cases. Explanation:

(?=( \S+)+) x^$#1 

Insert all of the powers of x, including x^1 but not x^0.

\b0x.\d+ 

Delete all powers of x with zero coefficients, but not a trailing 0 (yet).

\b1x x 

Delete a multiplier of 1 (but not a constant 1).

x.1 x 

Delete the ^1 of x^1.

 0 

Delete a constant 0 unless it is the only thing left.

 - - 

Delete the space before a -.

  + 

Change any remaining spaces to +s.

JavaScript (ES6), 107 106 bytes

a=>a.map(c=>c?s+=(c>0&&s?'+':_)+(!--e|c*c-1?c:c<0?'-':_)+(e?e-1?'x^'+e:'x':_):e--,e=a.length,_=s='')&&s||0 

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How?

The output is built by applying the following formulas to each coefficient c of the input array a[ ] while keeping track of the current exponent e.

1st formula: plus sign

If the coefficient is strictly positive and this is not the first term in the output expression, we append a +. Otherwise, we append nothing.

c > 0 && s ? '+' : _ 

2nd formula: minus sign and coefficient

If the exponent is zero or the absolute value of the coefficient is not equal to 1, we append the coefficient (which may include a leading -). Otherwise, we append either a - (if the coefficient is negative) or nothing.

!--e | c * c - 1 ? c : c < 0 ? '-' : _ 

3rd formula: variable and exponent

If the exponent is 0, we append nothing. If the exponent is 1, we append x. Otherwise, we append x^ followed by the exponent.

e ? e - 1 ? 'x^' + e : 'x' : _ 

Stax, 37 bytes

┴₧↕ê♦•Vªâÿσ9s╘dσ■à@@ⁿ■o─╦ñºº┌x╡ER▓ δ¿ 

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Here's the unpacked, ungolfed version.

r{ reverse the input and map using block ... |c skip this coefficient if it's falsy (zero) 0<.+-@ sign char; e.g. '+' _|aYv i!+ abs(coeff)!=1 || i>0 y$ str(abs(coeff)); e.g. '7' z "" ?+ if-then-else, concatenate; e.g. '+7' "x^`i" string template e.g. 'x^3' or 'x^0' iJ(T+ truncate string at i*i and trim. e.g. 'x^3' or '' mr$ re-reverse mapped array, and flatten to string c43=t if it starts with '+', drop the first character c0? if the result is blank, use 0 instead 

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Python 3, 279 277 258 251 bytes

k=str.replace def f(x): z=len(x) y='--'*(['-1']==[c for c in x if'0'!=c][:1]) for i,v in enumerate(x): p=str(z+~i) if v in'-1'and~i+z:y+='+x^'+p elif'0'!=v:y+='+'+v+'x^'+p return y and k(k(k(k(y[1:],'+-','-'),'^1',''),'x^0',''),'-+','-')or 0 

Takes input as a list of strings. This solution is not highly golfed yet. This basically works by replacing things to suit the output format, which highly increases the byte count.

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Special thanks to ovs and NK1406.

Pari/GP, 41 bytes

p->concat([c|c<-Vec(Str(Pol(p))),c!="*"]) 

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If an * between the coefficient and the variable is allowed:

Pari/GP, 3 bytes

Pol 

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APL (Dyalog Classic), 114 113 109 107 106 bytes

{{⍵≡'':⍕0⋄⍵↓⍨'+'=⊃⍵}∊⍵{{'1x'≡2↑1↓⍵:¯1⌽1↓1⌽⍵⋄⍵}('-0+'[1+×⍺]~⍕0),∊(U/⍕|⍺),(U←⍺≠0)/(⍵>⍳2)/¨'x'('^',⍕⍵)}¨⌽⍳⍴⍵} 

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-4 bytes thanks to @dzaima!

This can definitely be golfed down further. This requires ⎕IO←0

Pip, 78 bytes

(RV(B."x^"._MERVg)J'+)R[`\b0[^+]+``x.0|\^1|^\++|\++$``\b1x``\++-?`][xx'x_@v]|0 

Takes the coefficients as command-line arguments. Try it online!

Uses ME (map-enumerate) and J (join) to generate something of the form 0x^3+-1x^2+35x^1+0x^0, and then a bunch of regex replacements to transform this into the proper format.

APL (Dyalog Classic), 79 76 bytes

{'^$' '\b1x' '\+?¯' '^\+'⎕r('0x-',⊂⍬)∊⍕¨(0≠⍵)⌿'+',⍵,⊖⍪((⊢↓⍨2≤⊢)↑'x^',⍕)¨⍳≢⍵} 

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Python 3, 161 162 bytes

Fixed a bug thanks to ovs.

l=len lambda p:''.join(['+'*(i>0)*(c>0)+(str(c)[:-1],str(c))[abs(c)!=1or i==l(p)-1]+'x'*(i!=l(p)-1)+('^%d'%(l(p)+~i))*(i<l(p)-2)for i,c in enumerate(p)if c])or'0' 

Expanded:

l=len # Alias the len function since we use it a lot lambda p: ''.join([ # Join a list of strings '+'*(i>0)*(c>0) # Prepend a + if this isn't the first term and the coefficient is positive + (str(c)[:-1], str(c))[abs(c) != 1 or i == l(p) - 1] # If the coefficient is 1 and this isn't the last term, delete the '1' from the string representation, otherwise just use the string representation + 'x' * (i != l(p) - 1) # If this isn't the last term, append an x + ('^%d' % (l(p) + ~i)) * (i < l(p) - 2) # If this isn't one of the last two terms, append the exponent for i, c in enumerate(p) if c]) # Iterating over each coefficient with its index, discarding the term if the coefficient is zero or '0' # If all of the above resulted in an empty string, replace it with '0' 

C#, 237 bytes

c=>{var b=1>0;var r="";int l=c.Length;var p=!b;for(int i=0;i<l;i++){int n=c[i];int e=l-1-i;var o=p&&i>0&&n>0?"+":n==-1&&e!=0?"-":"";p=n!=0?b:p;r+=n==0?"":o+(e==0?$"{n}":e==1?$"{n}x":n==1||n==-1?$"x^{e}":$"{n}x^{e}");}return r==""?"0":r;} 

Clean, 172 bytes

import StdEnv,Text r=reverse @""="0" @a|a.[size a-1]<'0'=a+"1"=a ? -1="-" ?1="" ?a=a<+"" $l=join"-"(split"+-"(join"+"(r[?v+e\\v<-r l&e<-["","x":map((<+)"x^")[1..]]|v<>0]))) 

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Wolfram Language/Mathematica, 39 bytes

TraditionalForm@Expand@FromDigits[#,x]& 

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Turns out there is a built-in to get in in the right order.

Previous solution:

Wolfram Language/Mathematica, 93 bytes

StringReplace[StringRiffle[ToString/@InputForm/@MonomialList@FromDigits[#,x],"+"],"+-"->"-"]& 

At least to me, this is suprisingly long for a language designed for mathematical manipulation. It seems like Expand@FromDigits[#,x]& should work, but the default ordering for polynomials is the reverse of what the question requires, so some extra finagling is required.

Explanation

FromDigits[#,x] converts input list to polynomial (technically converts to a number in base x) MonomialList@ gets list of terms of polynomial InputForm/@ converts each term to the form a*x^n ToString/@ then to a string version of that StringRiffle[...,"+"] joins using +'s StringReplace[...,"+-"->"-"]& replaces +-'s with -'s 

Python3: 150 146 bytes

f=lambda l:''.join('+-'[a<0]+str(a)[a<0:5*((abs(a)!=1)|(1>i))]+'x^'[:i]+str(i)[:i-1]for i,a in zip(range(len(l)-1,-1,-1),l)if a).lstrip('+')or '0' 

(previous implementations):

f=lambda l: ''.join('+-'[a<0]+str(a)[a<0:5*((abs(a)!=1)|(1>i))]+'x^'[:i]+str(i)[:i-1] for i,a in zip(range(len(l)-1,-1,-1),l) if a).lstrip('+') or '0' 

You can try it online

Kudos to: @Benjamin

Perl 5 -a, 94 bytes

($_=shift@F)&&push@a,@a*!/-/>0&&'+',@F?s/\b1\b//r:$_,@F>0&&'x',@F>1&&'^'.@F while@F;say@a?@a:0 

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Retina 0.8.2, 113 bytes

\w+ a$'b$& a( [^b ]*)*?b(\d+) $2x^$#1 (\^1|x\^0)(?!\d) (?<= |-|^)(1(?=x)|0[^ -]*) ([ -])* $1 [ -]$|^  ^$ 0   + 

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I'm sure there is a lot to golf here...

Haskell, 166 163 bytes

g s|l<-length s,v:w:r<-id=<<["- +"!!(1+signum m):(id=<<[show$abs m|abs m>1||e==0]++["x"|e>0]++['^':show e|e>1])|(e,m)<-zip[l-1,l-2..]s,m/=0]=[v|v>'+']++w:r|1<3="0" 

Try it online! Example usage: g [0,-1,35,0] yields "-x^2+35x".

Previous 166 byte solution, which is slightly better readable:

0#n=show n m#n=id=<<[show n|n>1]++"x":['^':show m|m>1] m%0="" m%n|n<0='-':m#(-n)|1<3='+':m#n g s|l<-length s,v:m:r<-id=<<zipWith(%)[l-1,l-2..]s=[v|v>'+']++m:r|1<3="0" 

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Ruby, 111 bytes

->a{i=a.size;s=a.map{|x|i-=1;"%+d"%x+[?x,"x^#{i}",""][i<=>1]if x!=0}*'';s[0]?s.gsub(/(?<!\d)1(?=x)|^\+/,""):?0} 

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Solving this in Ruby turned out to be a bit frustrating, mainly due to the fact that unlike most languages, in Ruby (almost) everything is truthy, including 0-s and empty strings, so that even a simple check for zero becomes nowhere near as short as x?.

I played with various methods of constructing the string, and ultimately settled on a mix of several approaches:

• Terms with 0 coefficients are dropped using a simple conditional
• + and - signs are produced by formatting syntax with forced sign: %+d
• The correct form or x^i is selected using rocket operator indexing [...][i<=>1]
• Leading + and unnecessary 1-s are removed by regex replacements

Husk, 44 43 41 40 bytes

|s0Ψf¤|□ṁ`:'+f¹zμ+↓s²_&ε²¹↑□¹+"x^"s)¹m←ṡ 

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This feels a bit clunky; Husk is not optimized for string manipulation. I borrowed some ideas from the Stax answer.

Explanation

 Implicit input, say L = [2,-3,0,-1]. First we compute the exponents. ṡ Reversed indices: [4,3,2,1] m← Decrement each: [3,2,1,0] Then we format the individual terms of the polynomial. zμ...)¹ Zip with L using two-argument lambda: Arguments are coefficient and index, say C = -3 and I = 2. +"x^"s Convert I to string and concatenate to "x^": "x^2" ↑□¹ Take first I*I characters (relevant when I = 0 or I = 1): "x^2" _&ε²¹ Check if abs(C) <= 1 and I != 0, negate; returns -1 if true, 0 if false. ↓s² Convert C to string and drop that many elements (from the end, since negative). Result: "-3" The drop is relevant if C = 1 or C = -1. + Concatenate: "-3x^2" Result of zipping is ["2x^3","-3x^2","x","-1"] f¹ Keep those where the corresponding element of L is nonzero: ["2x^3","-3x^2","-1"] Next we join the terms with + and remove extraneous +s. ṁ Map and concatenate `:'+ appending '+': "2x^3+-3x^2+-1+" Ψf Adjacent filter: keep those chars A with right neighbor B ¤|□ where at least one of A or B is alphanumeric: "2x^3-3x^2-1" |s0 Finally, if the result is empty, return "0" instead. 

Perl 6, 97 bytes

{$!=+$_;.map({('+'x?($_&&$++&$_>0)~.substr(--$!&&2>.abs)~(<<''x>>[$!]//'x^'~$!))x?$_}).join||'0'} 

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Explanation:

$!=+$_; 

$! keeps track of the current exponent.

'+'x?($_&&$++&$_>0) 

Add + before positive coefficients, except if it's the first non-zero one. The $_&& short circuit makes sure that the anonymous state variable $ is only incremented for non-zero coefficients. The & junction is collapsed when coerced to Bool with ?.

.substr(--$!&&2>.abs) 

Decrement $!. Chop 1 or -1 coefficient unless it's constant.

<<''x>>[$!]//'x^'~$! 

Special-case linear and constant terms. Using the quote-protecting << >> construct is one byte shorter than the equivalent ('','x') or 2>$!??'x'x$!!!'x^'~$!.

x?$_ 

Hide zero terms, but always evaluate the preceding expression for the --$! side effect.

||'0' 

Return 0 if all coefficients are zero.

Java 8, 202 176 174 173 bytes

a->{String r="";int j=a.length;for(int i:a)r+=i==0*j--?"":"+"+i+(j<1?"":j<2?"x":"x^"+j);return r.isEmpty()?"0":r.substring(1).replace("+-","-").replaceAll("(\\D)1x","$1x");} 

• 26 bytes thanks to @Nevay.

Explanation:

Try it online.

a->{ // Method with String-array parameter and String return-type String r=""; // Result-String, starting empty int j=a.length; // Power-integer, starting at the size of the input-array for(int i:a) // Loop over the array r+=i==0 // If the current item is 0 *j--? // (And decrease `j` by 1 at the same time) "" // Append the result with nothing : // Else: "+" // Append the result with a "+", +i // and the current item, +(j<1? // +If `j` is 0: "" // Append nothing more :j<2? // Else-if `j` is 1: "x" // Append "x" : // Else: "x^"+j); // Append "x^" and `j` return r.isEmpty()? // If `r` is still empty "0" // Return "0" : // Else: r.substring(1) // Return the result minus the leading "+", .replace("+-","-") // and change all occurrences of "+-" to "-", .replaceAll("(\\D)1x","$1x");} // and all occurrences of "1x" to "x" 

Python, 165 bytes

lambda a:"".join([("+"if c>0 and i+1<len(a)else"")+(str(c)if i==0 or abs(c)!=1 else "")+{0:"",1:"x"}.get(i,"x^"+str(i))for i,c in enumerate(a[::-1])if c][::-1])or"0" 

PHP, 213 Bytes

$p=0;for($i=count($a=array_reverse(explode(',',trim($argv[1],'[]'))))-1;$i>=0;$i--)if(($b=(float)$a[$i])||(!$i&&!$p)){$k=abs($b);echo ($b<0?'-':($p?'+':'')).((($k!=1)||!$i)?$k:'').($i>1?'x^'.$i:($i?'x':''));$p=1;} 

Command line argument as requested by OP (single argument with brackets and commas).

Pretty print and some explanation:

$p = false; /* No part of the polynomial has yet been printed. */ for ($i = count($a = array_reverse(explode(',',trim($argv[1],'[]')))) - 1; $i >= 0; $i--) { $b = (float)$a[$i]; /* Cast to float to avoid -0 and numbers like 1.0 */ if (($b != 0) or (($i == 0) and !$p)) /* Print, if $b != 0 or the constant if there is no part until here. */ { $k = abs($b); echo ($b < 0 ? '-' : ( $p ? '+' : '')); /* Sign. The first sign is suppressed (if $p is false) if $b positive. */ echo ((($k != 1) || ($i == 0)) ? $k : ''); /* Coefficient */ echo ($i > 1 ? 'x^' . $i : (($i != 0) ? 'x' : '')); /* x^3, x^2, x, constant with empty string. */ $p = true; /* Part of the polynomial has been printed. */ } } 

PowerShell, 295 bytes

$c=$args[0] $p=$c.length-1 $i=0 $q="" while($p -ge 0){$t="";$e="";$d=$c[$i];switch($p){0{$t=""}1{$t="x"}Default{$t="x^";$e=$p}}if($d-eq 0){$t=""}elseif($d-eq 1){$t="+$t$e"}elseif($d-eq-1){$t="-$t$e"}elseif($d-lt 0 -or$i -eq 0){$t="$d$t$e"}else{$t="+$d$t$e"}$q+=$t;$i++;$p--}if($q -eq""){$q=0} $q