Roots of a function

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Task
Roots of a function
You are encouraged to solve this task according to the task description, using any language you may know.
Create a program that finds and outputs the roots of a given function, range and (if applicable) step width. The program should identify whether the root is exact or approximate.

For this example, use f(x)=x3-3x2+2x.

Contents

[edit] Ada

with Ada.Text_Io; use Ada.Text_Io;
 
procedure Roots_Of_Function is
package Real_Io is new Ada.Text_Io.Float_Io(Long_Float);
use Real_Io;
 
function F(X : Long_Float) return Long_Float is
begin
return (X**3 - 3.0*X*X + 2.0*X);
end F;
 
Step  : constant Long_Float := 1.0E-6;
Start : constant Long_Float := -1.0;
Stop  : constant Long_Float := 3.0;
Value : Long_Float := F(Start);
Sign  : Boolean := Value > 0.0;
X  : Long_Float := Start + Step;
 
begin
if Value = 0.0 then
Put("Root found at ");
Put(Item => Start, Fore => 1, Aft => 6, Exp => 0);
New_Line;
end if;
while X <= Stop loop
Value := F(X);
if (Value > 0.0) /= Sign then
Put("Root found near ");
Put(Item => X, Fore => 1, Aft => 6, Exp => 0);
New_Line;
elsif Value = 0.0 then
Put("Root found at ");
Put(Item => X, Fore => 1, Aft => 6, Exp => 0);
New_Line;
end if;
Sign := Value > 0.0;
X := X + Step;
end loop;
end Roots_Of_Function;

[edit] ALGOL 68

Works with: ALGOL 68 version Revision 1 - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

Finding 3 roots using the secant method:

MODE DBL = LONG REAL;
FORMAT dbl = $g(-long real width, long real width-6, -2)$;
 
MODE XY = STRUCT(DBL x, y);
FORMAT xy root = $f(dbl)" ("b("Exactly", "Approximately")")"$;
 
MODE DBLOPT = UNION(DBL, VOID);
MODE XYRES = UNION(XY, VOID);
 
PROC find root = (PROC (DBL)DBL f, DBLOPT in x1, in x2, in x error, in y error)XYRES:(
INT limit = ENTIER (long real width / log(2)); # worst case of a binary search) #
DBL x1 := (in x1|(DBL x1):x1|-5.0), # if x1 is EMPTY then -5.0 #
x2 := (in x2|(DBL x2):x2|+5.0),
x error := (in x error|(DBL x error):x error|small real),
y error := (in y error|(DBL y error):y error|small real);
DBL y1 := f(x1), y2;
DBL dx := x1 - x2, dy;
 
IF y1 = 0 THEN
XY(x1, y1) # we already have a solution! #
ELSE
FOR i WHILE
y2 := f(x2);
IF y2 = 0 THEN stop iteration FI;
IF i = limit THEN value error FI;
IF y1 = y2 THEN value error FI;
dy := y1 - y2;
dx := dx / dy * y2;
x1 := x2; y1 := y2; # retain for next iteration #
x2 -:= dx;
# WHILE # ABS dx > x error AND ABS dy > y error DO
SKIP
OD;
stop iteration:
XY(x2, y2) EXIT
value error:
EMPTY
FI
);
 
PROC f = (DBL x)DBL: x UP 3 - LONG 3.1 * x UP 2 + LONG 2.0 * x;
 
DBL first root, second root, third root;
 
XYRES first result = find root(f, LENG -1.0, LENG 3.0, EMPTY, EMPTY);
CASE first result IN
(XY first result): (
printf(($"1st root found at x = "f(xy root)l$, x OF first result, y OF first result=0));
first root := x OF first result
)
OUT printf($"No first root found"l$); stop
ESAC;
 
XYRES second result = find root( (DBL x)DBL: f(x) / (x - first root), EMPTY, EMPTY, EMPTY, EMPTY);
CASE second result IN
(XY second result): (
printf(($"2nd root found at x = "f(xy root)l$, x OF second result, y OF second result=0));
second root := x OF second result
)
OUT printf($"No second root found"l$); stop
ESAC;
 
XYRES third result = find root( (DBL x)DBL: f(x) / (x - first root) / ( x - second root ), EMPTY, EMPTY, EMPTY, EMPTY);
CASE third result IN
(XY third result): (
printf(($"3rd root found at x = "f(xy root)l$, x OF third result, y OF third result=0));
third root := x OF third result
)
OUT printf($"No third root found"l$); stop
ESAC

Output:

1st root found at x =  9.1557112297752398099031e-1 (Approximately)
2nd root found at x =  2.1844288770224760190097e 0 (Approximately)
3rd root found at x =  0.0000000000000000000000e 0 (Exactly)

[edit] AutoHotkey

Poly(x) is a test function of one variable, here we are searching for its roots:

  • roots() searches for intervals within given limits, shifted by a given “step”, where our function has different signs at the endpoints.
  • Having found such an interval, the root() function searches for a value where our function is 0, within a given tolerance.
  • It also sets ErrorLevel to info about the root found.

discussion

MsgBox % roots("poly", -0.99, 2, 0.1, 1.0e-5)
MsgBox % roots("poly", -1, 3, 0.1, 1.0e-5)
 
roots(f,x1,x2,step,tol) { ; search for roots in intervals of length "step", within tolerance "tol"
x := x1, y := %f%(x), s := (y>0)-(y<0)
Loop % ceil((x2-x1)/step) {
x += step, y := %f%(x), t := (y>0)-(y<0)
If (s=0 || s!=t)
res .= root(f, x-step, x, tol) " [" ErrorLevel "]`n"
s := t
}
Sort res, UN ; remove duplicate endpoints
Return res
}
 
root(f,x1,x2,d) { ; find x in [x1,x2]: f(x)=0 within tolerance d, by bisection
If (!y1 := %f%(x1))
Return x1, ErrorLevel := "Exact"
If (!y2 := %f%(x2))
Return x2, ErrorLevel := "Exact"
If (y1*y2>0)
Return "", ErrorLevel := "Need different sign ends!"
Loop {
x := (x2+x1)/2, y := %f%(x)
If (y = 0 || x2-x1 < d)
Return x, ErrorLevel := y ? "Approximate" : "Exact"
If ((y>0) = (y1>0))
x1 := x, y1 := y
Else
x2 := x, y2 := y
}
}
 
poly(x) {
Return ((x-3)*x+2)*x
}

[edit] Axiom

Using a polynomial solver:

expr := x^3-3*x^2+2*x
solve(expr,x)

Output:

  (1)  [x= 2,x= 1,x= 0]
Type: List(Equation(Fraction(Polynomial(Integer))))

Using the secant method in the interpreter:

digits(30)
secant(eq: Equation Expression Float, binding: SegmentBinding(Float)):Float ==
eps := 1.0e-30
expr := lhs eq - rhs eq
x := variable binding
seg := segment binding
x1 := lo seg
x2 := hi seg
fx1 := eval(expr, x=x1)::Float
abs(fx1)<eps => return x1
for i in 1..100 repeat
fx2 := eval(expr, x=x2)::Float
abs(fx2)<eps => return x2
(x1, fx1, x2) := (x2, fx2, x2 - fx2 * (x2 - x1) / (fx2 - fx1))
error "Function not converging."

The example can now be called using:

secant(expr=0,x=-0.5..0.5)

[edit] BBC BASIC

      function$ = "x^3-3*x^2+2*x"
rangemin = -1
rangemax = 3
stepsize = 0.001
accuracy = 1E-8
PROCroots(function$, rangemin, rangemax, stepsize, accuracy)
END
 
DEF PROCroots(func$, min, max, inc, eps)
LOCAL x, sign%, oldsign%
oldsign% = 0
FOR x = min TO max STEP inc
sign% = SGN(EVAL(func$))
IF sign% = 0 THEN
PRINT "Root found at x = "; x
sign% = -oldsign%
ELSE IF sign% <> oldsign% AND oldsign% <> 0 THEN
IF inc < eps THEN
PRINT "Root found near x = "; x
ELSE
PROCroots(func$, x-inc, x+inc/8, inc/8, eps)
ENDIF
ENDIF
ENDIF
oldsign% = sign%
NEXT x
ENDPROC

Output:

Root found near x = 2.29204307E-9
Root found near x = 1
Root found at x = 2

[edit] C

Step & check until close, then use Secant method.

#include <math.h>
#include <stdio.h>
 
double f(double x)
{
return x*x*x-3.0*x*x +2.0*x;
}
 
double secant( double xA, double xB, double(*f)(double) )
{
double e = 1.0e-12;
double fA, fB;
double d;
int i;
int limit = 50;
 
fA=(*f)(xA);
for (i=0; i<limit; i++) {
fB=(*f)(xB);
d = (xB - xA) / (fB - fA) * fB;
if (fabs(d) < e)
break;
xA = xB;
fA = fB;
xB -= d;
}
if (i==limit) {
printf("Function is not converging near (%7.4f,%7.4f).\n", xA,xB);
return -99.0;
}
return xB;
}
 
int main(int argc, char *argv[])
{
double step = 1.0e-2;
double e = 1.0e-12;
double x = -1.032; // just so we use secant method
double xx, value;
 
int s = (f(x)> 0.0);
 
while (x < 3.0) {
value = f(x);
if (fabs(value) < e) {
printf("Root found at x= %12.9f\n", x);
s = (f(x+.0001)>0.0);
}
else if ((value > 0.0) != s) {
xx = secant(x-step, x,&f);
if (xx != -99.0) // -99 meaning secand method failed
printf("Root found at x= %12.9f\n", xx);
else
printf("Root found near x= %7.4f\n", x);
s = (f(x+.0001)>0.0);
}
x += step;
}
return 0;
}

[edit] C++

#include <iostream>
 
double f(double x)
{
return (x*x*x - 3*x*x + 2*x);
}
 
int main()
{
double step = 0.001; // Smaller step values produce more accurate and precise results
double start = -1;
double stop = 3;
double value = f(start);
double sign = (value > 0);
 
// Check for root at start
if ( 0 == value )
std::cout << "Root found at " << start << std::endl;
 
for( double x = start + step;
x <= stop;
x += step )
{
value = f(x);
 
if ( ( value > 0 ) != sign )
// We passed a root
std::cout << "Root found near " << x << std::endl;
else if ( 0 == value )
// We hit a root
std::cout << "Root found at " << x << std::endl;
 
// Update our sign
sign = ( value > 0 );
}
}

[edit] Clojure

Translation of: Haskell
 
 
(defn findRoots [f start stop step eps]
(filter #(-> (f %) Math/abs (< eps)) (range start stop step)))
 
> (findRoots #(+ (* % % %) (* -3 % %) (* 2 %)) -1.0 3.0 0.0001 0.00000001)
(-9.381755897326649E-14 0.9999999999998124 1.9999999999997022)

[edit] CoffeeScript

Translation of: Python
 
print_roots = (f, begin, end, step) ->
# Print approximate roots of f between x=begin and x=end,
# using sign changes as an indicator that a root has been
# encountered.
x = begin
y = f(x)
last_y = y
 
cross_x_axis = ->
(last_y < 0 and y > 0) or (last_y > 0 and y < 0)
 
console.log '-----'
while x <= end
y = f(x)
if y == 0
console.log "Root found at", x
else if cross_x_axis()
console.log "Root found near", x
x += step
last_y = y
 
do ->
# Smaller steps produce more accurate/precise results in general,
# but for many functions we'll never get exact roots, either due
# to imperfect binary representation or irrational roots.
step = 1 / 256
 
f1 = (x) -> x*x*x - 3*x*x + 2*x
print_roots f1, -1, 5, step
f2 = (x) -> x*x - 4*x + 3
print_roots f2, -1, 5, step
f3 = (x) -> x - 1.5
print_roots f3, 0, 4, step
f4 = (x) -> x*x - 2
print_roots f4, -2, 2, step
 

output

 
> coffee roots.coffee
-----
Root found at 0
Root found at 1
Root found at 2
-----
Root found at 1
Root found at 3
-----
Root found at 1.5
-----
Root found near -1.4140625
Root found near 1.41796875
 

[edit] Common Lisp

Translation of: Perl

find-roots prints roots (and values near roots) and returns a list of root designators, each of which is either a number n, in which case (zerop (funcall function n)) is true, or a cons whose car and cdr are such that the sign of function at car and cdr changes.

(defun find-roots (function start end &optional (step 0.0001))
(let* ((roots '())
(value (funcall function start))
(plusp (plusp value)))
(when (zerop value)
(format t "~&Root found at ~W." start))
(do ((x (+ start step) (+ x step)))
((> x end) (nreverse roots))
(setf value (funcall function x))
(cond
((zerop value)
(format t "~&Root found at ~w." x)
(push x roots))
((not (eql plusp (plusp value)))
(format t "~&Root found near ~w." x)
(push (cons (- x step) x) roots)))
(setf plusp (plusp value)))))
> (find-roots #'(lambda (x) (+ (* x x x) (* -3 x x) (* 2 x))) -1 3)
Root found near 5.3588345E-5.
Root found near 1.0000072.
Root found near 2.000073.
((-4.6411653E-5 . 5.3588345E-5)
 (0.99990714 . 1.0000072)
 (1.9999729 . 2.000073))

[edit] D

import std.stdio, std.math, std.algorithm;
 
bool nearZero(T)(in T a, in T b = T.epsilon * 4) pure nothrow {
return abs(a) <= b;
}
 
T[] findRoot(T)(immutable T function(in T) pure nothrow fi,
in T start, in T end, in T step=T(0.001L),
T tolerance = T(1e-4L)) {
if (step.nearZero)
writefln("WARNING: step size may be too small.");
 
/// Search root by simple bisection.
T searchRoot(T a, T b) pure nothrow {
T root;
int limit = 49;
T gap = b - a;
 
while (!nearZero(gap) && limit--) {
if (fi(a).nearZero)
return a;
if (fi(b).nearZero)
return b;
root = (b + a) / 2.0L;
if (fi(root).nearZero)
return root;
((fi(a) * fi(root) < 0) ? b : a) = root;
gap = b - a;
}
 
return root;
}
 
immutable dir = T(end > start ? 1.0 : -1.0);
immutable step2 = (end > start) ? abs(step) : -abs(step);
T[T] result;
for (T x = start; (x * dir) <= (end * dir); x += step2)
if (fi(x) * fi(x + step2) <= 0) {
immutable T r = searchRoot(x, x + step2);
result[r] = fi(r);
}
 
return result.keys.sort().release;
}
 
void report(T)(in T[] r, immutable T function(in T) pure f,
in T tolerance = T(1e-4L)) {
if (r.length) {
writefln("Root found (tolerance = %1.4g):", tolerance);
 
foreach (const x; r) {
immutable T y = f(x);
 
if (nearZero(y))
writefln("... EXACTLY at %+1.20f, f(x) = %+1.4g",x,y);
else if (nearZero(y, tolerance))
writefln(".... MAY-BE at %+1.20f, f(x) = %+1.4g",x,y);
else
writefln("Verify needed, f(%1.4g) = " ~
"%1.4g > tolerance in magnitude", x, y);
}
} else
writefln("No root found.");
}
 
void main() {
static real f(in real x) pure nothrow {
return x ^^ 3 - (3 * x ^^ 2) + 2 * x;
}
 
findRoot(&f, -1.0L, 3.0L, 0.001L).report(&f);
}
Output:
Root found (tolerance = 0.0001):
.... MAY-BE at -0.00000000000000000080, f(x) = -1.603e-18
... EXACTLY at +1.00000000000000000020, f(x) = -2.168e-19
.... MAY-BE at +1.99999999999999999950, f(x) = -8.674e-19

NB: smallest increment for real type in D is real.epsilon = 1.0842e-19.

[edit] DWScript

Translation of: C
type TFunc = function (x : Float) : Float;
 
function f(x : Float) : Float;
begin
Result := x*x*x-3.0*x*x +2.0*x;
end;
 
const e = 1.0e-12;
 
function Secant(xA, xB : Float; f : TFunc) : Float;
const
limit = 50;
var
fA, fB : Float;
d : Float;
i : Integer;
begin
fA := f(xA);
for i := 0 to limit do begin
fB := f(xB);
d := (xB-xA)/(fB-fA)*fB;
if Abs(d) < e then
Exit(xB);
xA := xB;
fA := fB;
xB -= d;
end;
PrintLn(Format('Function is not converging near (%7.4f,%7.4f).', [xA, xB]));
Result := -99.0;
end;
 
const fstep = 1.0e-2;
 
var x := -1.032; // just so we use secant method
var xx, value : Float;
var s := f(x)>0.0;
 
while (x < 3.0) do begin
value := f(x);
if Abs(value)<e then begin
PrintLn(Format("Root found at x= %12.9f", [x]));
s := (f(x+0.0001)>0.0);
end else if (value>0.0) <> s then begin
xx := Secant(x-fstep, x, f);
if xx <> -99.0 then // -99 meaning secand method failed
PrintLn(Format('Root found at x = %12.9f', [xx]))
else PrintLn(Format('Root found near x= %7.4f', [xx]));
s := (f(x+0.0001)>0.0);
end;
x += fstep;
end;

[edit] Erlang

% Implemented by Arjun Sunel
-module(roots).
-export([main/0]).
main() ->
F = fun(X)->X*X*X - 3*X*X + 2*X end,
Step = 0.001, % Using smaller steps will provide more accurate results
Start = -1,
Stop = 3,
Sign = F(Start) > 0,
X = Start,
while(X, Step, Start, Stop, Sign,F).
 
while(X, Step, Start, Stop, Sign,F) ->
Value = F(X),
if
Value == 0 -> % We hit a root
io:format("Root found at ~p~n",[X]),
while(X+Step, Step, Start, Stop, Value > 0,F);
 
(Value < 0) == Sign -> % We passed a root
io:format("Root found near ~p~n",[X]),
while(X+Step , Step, Start, Stop, Value > 0,F);
 
X > Stop ->
io:format("") ;
true ->
while(X+Step, Step, Start, Stop, Value > 0,F)
end.
 
Output:
Root found near 8.81239525796218e-16
Root found near 1.0000000000000016
Root found near 2.0009999999998915
ok

[edit] Fortran

Works with: Fortran version 90 and later
PROGRAM ROOTS_OF_A_FUNCTION
 
IMPLICIT NONE
 
INTEGER, PARAMETER :: dp = SELECTED_REAL_KIND(15)
REAL(dp) :: f, e, x, step, value
LOGICAL :: s
 
f(x) = x*x*x - 3.0_dp*x*x + 2.0_dp*x
 
x = -1.0_dp ; step = 1.0e-6_dp ; e = 1.0e-9_dp
 
s = (f(x) > 0)
DO WHILE (x < 3.0)
value = f(x)
IF(ABS(value) < e) THEN
WRITE(*,"(A,F12.9)") "Root found at x =", x
s = .NOT. s
ELSE IF ((value > 0) .NEQV. s) THEN
WRITE(*,"(A,F12.9)") "Root found near x = ", x
s = .NOT. s
END IF
x = x + step
END DO
 
END PROGRAM ROOTS_OF_A_FUNCTION

The following approach uses the Secant Method to numerically find one root. Which root is found will depend on the start values x1 and x2 and if these are far from a root this method may not converge.

INTEGER, PARAMETER :: dp = SELECTED_REAL_KIND(15)
INTEGER :: i=1, limit=100
REAL(dp) :: d, e, f, x, x1, x2
 
f(x) = x*x*x - 3.0_dp*x*x + 2.0_dp*x
 
x1 = -1.0_dp ; x2 = 3.0_dp ; e = 1.0e-15_dp
 
DO
IF (i > limit) THEN
WRITE(*,*) "Function not converging"
EXIT
END IF
d = (x2 - x1) / (f(x2) - f(x1)) * f(x2)
IF (ABS(d) < e) THEN
WRITE(*,"(A,F18.15)") "Root found at x = ", x2
EXIT
END IF
x1 = x2
x2 = x2 - d
i = i + 1
END DO

[edit] Go

Secant method. No error checking.

package main
 
import (
"fmt"
"math"
)
 
func f(x float64) float64 {
return x*x*x - 3*x*x + 2*x
}
 
var (
lower float64 = -.5
upper float64 = 2.6
step float64 = 1
)
 
func main() {
for x0, x1 := lower, lower+step; x0 < upper; x0, x1 = x1, x1+step {
if x1 > upper {
x1 = upper
}
r, status := secant(x0, x1)
if status != "" && r >= x0 && r < x1 {
fmt.Printf("  %6.3f %s\n", r, status)
}
}
}
 
func secant(x0, x1 float64) (float64, string) {
var f0 float64
f1 := f(x0)
for i := 0; i < 100; i++ {
f0, f1 = f1, f(x1)
switch {
case f1 == 0:
return x1, "exact"
case math.Abs(x1-x0) < 1e-6:
return x1, "approximate"
}
x0, x1 = x1, x1-f1*(x1-x0)/(f1-f0)
}
return 0, ""
}

Output:

   0.000 approximate
   1.000 exact
   2.000 approximate

[edit] Haskell

f x = x^3-3*x^2+2*x
 
findRoots start stop step eps =
[x | x <- [start, start+step .. stop], abs (f x) < eps]

Executed in GHCi:

*Main> findRoots (-1.0) 3.0 0.0001 0.000000001
[-9.381755897326649e-14,0.9999999999998124,1.9999999999997022]

Or using package hmatrix from HackageDB.

import Numeric.GSL.Polynomials
import Data.Complex
 
*Main> mapM_ print $ polySolve [0,2,-3,1]
(-5.421010862427522e-20) :+ 0.0
2.000000000000001 :+ 0.0
0.9999999999999996 :+ 0.0

No complex roots, so:

*Main> mapM_ (print.realPart) $ polySolve [0,2,-3,1]
-5.421010862427522e-20
2.000000000000001
0.9999999999999996

[edit] HicEst

HicEst's SOLVE function employs the Levenberg-Marquardt method:

OPEN(FIle='test.txt')
 
1 DLG(NameEdit=x0, DNum=3)
 
x = x0
chi2 = SOLVE(NUL=x^3 - 3*x^2 + 2*x, Unknown=x, I=iterations, NumDiff=1E-15)
EDIT(Text='approximate exact ', Word=(chi2 == 0), Parse=solution)
 
WRITE(FIle='test.txt', LENgth=6, Name) x0, x, solution, chi2, iterations
GOTO 1
x0=0.5; x=1; solution=exact; chi2=79E-32 iterations=65;
x0=0.4; x=2E-162 solution=exact; chi2=0; iterations=1E4;
x0=0.45; x=1; solution=exact; chi2=79E-32 iterations=67;
x0=0.42; x=2E-162 solution=exact; chi2=0; iterations=1E4;
x0=1.5; x=1.5; solution=approximate; chi2=0.1406; iterations=14:
x0=1.54; x=1; solution=exact; chi2=44E-32 iterations=63;
x0=1.55; x=2; solution=exact; chi2=79E-32 iterations=55;
x0=1E10; x=2; solution=exact; chi2=18E-31 iterations=511;
x0=-1E10; x=0; solution=exact; chi2=0; iterations=1E4;

[edit] J

J has builtin a root-finding operator, p., whose input is the coeffiecients of the polynomial (where the exponent of the indeterminate variable matches the index of the coefficient: 0 1 2 would be 0 + x + (2 times x squared)). Hence:

   1{::p.  0 2 _3 1    
2 1 0

We can determine whether the roots are exact or approximate by evaluating the polynomial at the candidate roots, and testing for zero:

   (0=]p.1{::p.) 0 2 _3 1 
1 1 1

As you can see, p. is also the operator which evaluates polynomials. This is not a coincidence.

That said, we could also implement the technique used by most others here. Specifically: we can implement the function as a black box and check every 1 millionth of a unit between minus one and three, and we can test that result for exactness.

   blackbox=: 0 2 _3 1&p.
(#~ (=<./)@:|@blackbox) i.&.(1e6&*)&.(1&+) 3
0 1 2
0=blackbox 0 1 2
1 1 1

Here, we see that each of the results (0, 1 and 2) are as accurate as we expect our computer arithmetic to be. (The = returns 1 where paired values are equal and 0 where they are not equal).

[edit] Java

public class Roots {
public interface Function {
public double f(double x);
}
 
private static int sign(double x) {
return (x < 0.0) ? -1 : (x > 0.0) ? 1 : 0;
}
 
public static void printRoots(Function f, double lowerBound,
double upperBound, double step) {
double x = lowerBound, ox = x;
double y = f.f(x), oy = y;
int s = sign(y), os = s;
 
for (; x <= upperBound ; x += step) {
s = sign(y = f.f(x));
if (s == 0) {
System.out.println(x);
} else if (s != os) {
double dx = x - ox;
double dy = y - oy;
double cx = x - dx * (y / dy);
System.out.println("~" + cx);
}
ox = x; oy = y; os = s;
}
}
 
public static void main(String[] args) {
Function poly = new Function () {
public double f(double x) {
return x*x*x - 3*x*x + 2*x;
}
};
printRoots(poly, -1.0, 4, 0.002);
}
}

Produces this output:

~2.616794878713638E-18
~1.0000000000000002
~2.000000000000001

[edit] Liberty BASIC

'   Finds and output the roots of a given function f(x),
' within a range of x values.
 
' [RC]Roots of an function
 
mainwin 80 12
 
xMin =-1
xMax = 3
y =f( xMin) ' Since Liberty BASIC has an 'eval(' function the fn
' and limits would be better entered via 'input'.
LastY =y
 
eps =1E-12 ' closeness acceptable
 
bigH=0.01
 
print
print " Checking for roots of x^3 -3 *x^2 +2 *x =0 over range -1 to +3"
print
 
x=xMin: dx = bigH
do
x=x+dx
y = f(x)
'print x, dx, y
if y*LastY <0 then 'there is a root, should drill deeper
if dx < eps then 'we are close enough
print " Just crossed axis, solution f( x) ="; y; " at x ="; using( "#.#####", x)
LastY = y
dx = bigH 'after closing on root, continue with big step
else
x=x-dx 'step back
dx = dx/10 'repeat with smaller step
end if
end if
loop while x<xMax
 
print
print " Finished checking in range specified."
 
end
 
function f( x)
f =x^3 -3 *x^2 +2 *x
end function

[edit] Maple

f := x^3-3*x^2+2*x;
roots(f,x);

outputs:

[[0, 1], [1, 1], [2, 1]]

which means there are three roots. Each root is named as a pair where the first element is the value (0, 1, and 2), the second one the multiplicity (=1 for each means none of the three are degenerate).

By itself (i.e. unless specifically asked to do so), Maple will only perform exact (symbolic) operations and not attempt to do any kind of numerical approximation.

[edit] Mathematica

There are multiple obvious ways to do this in Mathematica.

[edit] Solve

This requires a full equation and will perform symbolic operations only:

Solve[x^3-3*x^2+2*x==0,x]

Output

 {{x->0},{x->1},{x->2}}

[edit] NSolve

This requires merely the polynomial and will perform numerical operations if needed:

 NSolve[x^3 - 3*x^2 + 2*x , x]

Output

 {{x->0.},{x->1.},{x->2.}}

(note that the results here are floats)

[edit] FindRoot

This will numerically try to find one(!) local root from a given starting point:

FindRoot[x^3 - 3*x^2 + 2*x , {x, 1.5}]

Output

 {x->0.}

From a different start point:

FindRoot[x^3 - 3*x^2 + 2*x , {x, 1.1}]

Output

{x->1.}

(note that there is no guarantee which one is found).

[edit] FindInstance

This finds a value (optionally out of a given domain) for the given variable (or a set of values for a set of given variables) that satisfy a given equality or inequality:

 FindInstance[x^3 - 3*x^2 + 2*x == 0, x]

Output

{{x->0}}

[edit] Reduce

This will (symbolically) reduce a given expression to the simplest possible form, solving equations and performing substitutions in the process:

Reduce[x^3 - 3*x^2 + 2*x == 0, x]
 x==0||x==1||x==2

(note that this doesn't yield a "solution" but a different expression that expresses the same thing as the original)

[edit] Maxima

e: x^3 - 3*x^2 + 2*x$
 
/* Number of roots in a real interval, using Sturm sequences */
nroots(e, -10, 10);
3
 
solve(e, x);
[x=1, x=2, x=0]
 
/* 'solve sets the system variable 'multiplicities */
 
solve(x^4 - 2*x^3 + 2*x - 1, x);
[x=-1, x=1]
 
multiplicities;
[1, 3]
 
/* Rational approximation of roots using Sturm sequences and bisection */
 
realroots(e);
[x=1, x=2, x=0]
 
/* 'realroots also sets the system variable 'multiplicities */
 
multiplicities;
[1, 1, 1]
 
/* Numerical root using Brent's method (here with another equation) */
 
find_root(sin(t) - 1/2, t, 0, %pi/2);
0.5235987755983
 
fpprec: 60$
 
bf_find_root(sin(t) - 1/2, t, 0, %pi/2);
5.23598775598298873077107230546583814032861566562517636829158b-1
 
/* Numerical root using Newton's method */
 
load(newton1)$
newton(e, x, 1.1, 1e-6);
1.000000017531147
 
/* For polynomials, Jenkins–Traub algorithm */
 
allroots(x^3 + x + 1);
[x=1.161541399997252*%i+0.34116390191401,
x=0.34116390191401-1.161541399997252*%i,
x=-0.68232780382802]
 
bfallroots(x^3 + x + 1);
[x=1.16154139999725193608791768724717407484314725802151429063617b0*%i + 3.41163901914009663684741869855524128445594290948999288901864b-1,
x=3.41163901914009663684741869855524128445594290948999288901864b-1 - 1.16154139999725193608791768724717407484314725802151429063617b0*%i,
x=-6.82327803828019327369483739711048256891188581897998577803729b-1]

[edit] Objeck

Translation of: C++
 
bundle Default {
class Roots {
function : f(x : Float) ~ Float
{
return (x*x*x - 3.0*x*x + 2.0*x);
}
 
function : Main(args : String[]) ~ Nil
{
step := 0.001;
start := -1.0;
stop := 3.0;
value := f(start);
sign := (value > 0);
 
if(0.0 = value) {
start->PrintLine();
};
 
for(x := start + step; x <= stop; x += step;) {
value := f(x);
 
if((value > 0) <> sign) {
IO.Console->Instance()->Print("~")->PrintLine(x);
}
else if(0 = value) {
IO.Console->Instance()->Print("~")->PrintLine(x);
};
 
sign := (value > 0);
};
}
}
}
 

[edit] OCaml

A general root finder using the False Position (Regula Falsi) method, which will find all simple roots given a small step size.

let bracket u v =
((u > 0.0) && (v < 0.0)) || ((u < 0.0) && (v > 0.0));;
 
let xtol a b = (a = b);; (* or use |a-b| < epsilon *)
 
let rec regula_falsi a b fa fb f =
if xtol a b then (a, fa) else
let c = (fb*.a -. fa*.b) /. (fb -. fa) in
let fc = f c in
if fc = 0.0 then (c, fc) else
if bracket fa fc then
regula_falsi a c fa fc f
else
regula_falsi c b fc fb f;;
 
let search lo hi step f =
let rec next x fx =
if x > hi then [] else
let y = x +. step in
let fy = f y in
if fx = 0.0 then
(x,fx) :: next y fy
else if bracket fx fy then
(regula_falsi x y fx fy f) :: next y fy
else
next y fy in
next lo (f lo);;
 
let showroot (x,fx) =
Printf.printf "f(%.17f) = %.17f [%s]\n"
x fx (if fx = 0.0 then "exact" else "approx") in
let f x = ((x -. 3.0)*.x +. 2.0)*.x in
List.iter showroot (search (-5.0) 5.0 0.1 f);;

Output:

f(0.00000000000000000) = 0.00000000000000000 [exact]
f(1.00000000000000022) = 0.00000000000000000 [exact]
f(1.99999999999999978) = 0.00000000000000000 [exact]

Note these roots are exact solutions with floating-point calculation.

[edit] Octave

If the equation is a polynomial, we can put the coefficients in a vector and use roots:

a = [ 1, -3, 2, 0 ];
r = roots(a);
% let's print it
for i = 1:3
n = polyval(a, r(i));
printf("x%d = %f (%f", i, r(i), n);
if (n != 0.0)
printf(" not");
endif
printf(" exact)\n");
endfor

Otherwise we can program our (simple) method:

Translation of: Python
function y = f(x)
y = x.^3 -3.*x.^2 + 2.*x;
endfunction
 
step = 0.001;
tol = 10 .* eps;
start = -1;
stop = 3;
se = sign(f(start));
 
x = start;
while (x <= stop)
v = f(x);
if ( (v < tol) && (v > -tol) )
printf("root at %f\n", x);
elseif ( sign(v) != se )
printf("root near %f\n", x);
endif
se = sign(v);
x = x + step;
endwhile

[edit] PARI/GP

[edit] Gourdon–Schönhage algorithm

polroots(x^3-3*x^2+2*x)

[edit] Newton's method

This uses a modified version of the Newton–Raphson method.

polroots(x^3-3*x^2+2*x,1)

[edit] Brent's method

solve(x=-.5,.5,x^3-3*x^2+2*x)
solve(x=.5,1.5,x^3-3*x^2+2*x)
solve(x=1.5,2.5,x^3-3*x^2+2*x)

[edit] Factorization to linear factors

findRoots(P)={
my(f=factor(P),t);
for(i=1,#f[,1],
if(poldegree(f[i,1]) == 1,
for(j=1,f[i,2],
print(-polcoeff(f[i,1], 0), " (exact)")
)
);
if(poldegree(f[i,1]) > 1,
t=polroots(f[i,1]);
for(j=1,#t,
for(k=1,f[i,2],
print(if(imag(t[j]) == 0.,real(t[j]),t[j]), " (approximate)")
)
)
)
)
};
findRoots(x^3-3*x^2+2*x)

[edit] Factorization to quadratic factors

Of course this process could be continued to degrees 3 and 4 with sufficient additional work.

findRoots(P)={
my(f=factor(P),t);
for(i=1,#f[,1],
if(poldegree(f[i,1]) == 1,
for(j=1,f[i,2],
print(-polcoeff(f[i,1], 0), " (exact)")
)
);
if(poldegree(f[i,1]) == 2,
t=solveQuadratic(polcoeff(f[i,1],2),polcoeff(f[i,1],1),polcoeff(f[i,1],0));
for(j=1,f[i,2],
print(t[1]" (exact)\n"t[2]" (exact)")
)
);
if(poldegree(f[i,1]) > 2,
t=polroots(f[i,1]);
for(j=1,#t,
for(k=1,f[i,2],
print(if(imag(t[j]) == 0.,real(t[j]),t[j]), " (approximate)")
)
)
)
)
};
solveQuadratic(a,b,c)={
my(t=-b/2/a,s=b^2/4/a^2-c/a,inner=core(numerator(s))/core(denominator(s)),outer=sqrtint(s/inner));
if(inner < 0,
outer *= I;
inner *= -1
);
s=if(inner == 1,
outer
,
if(outer == 1,
Str("sqrt(", inner, ")")
,
Str(outer, " * sqrt(", inner, ")")
)
);
if (t,
[Str(t, " + ", s), Str(t, " - ", s)]
,
[s, Str("-", s)]
)
};
findRoots(x^3-3*x^2+2*x)

[edit] Pascal

Translation of: Fortran
Program RootsFunction;
 
var
e, x, step, value: double;
s: boolean;
i, limit: integer;
x1, x2, d: double;
 
function f(const x: double): double;
begin
f := x*x*x - 3*x*x + 2*x;
end;
 
begin
x := -1;
step := 1.0e-6;
e := 1.0e-9;
s := (f(x) > 0);
 
writeln('Version 1: simply stepping x:');
while x < 3.0 do
begin
value := f(x);
if abs(value) < e then
begin
writeln ('root found at x = ', x);
s := not s;
end
else if ((value > 0) <> s) then
begin
writeln ('root found at x = ', x);
s := not s;
end;
x := x + step;
end;
 
writeln('Version 2: secant method:');
x1 := -1.0;
x2 := 3.0;
e := 1.0e-15;
i := 1;
limit := 300;
while true do
begin
if i > limit then
begin
writeln('Error: function not converging');
exit;
end;
d := (x2 - x1) / (f(x2) - f(x1)) * f(x2);
if abs(d) < e then
begin
if d = 0 then
write('Exact ')
else
write('Approximate ');
writeln('root found at x = ', x2);
exit;
end;
x1 := x2;
x2 := x2 - d;
i := i + 1;
end;
end.
 

Output:

Version 1: simply stepping x:
root found at x =  7.91830063542152E-012
root found at x =  1.00000000001584E+000
root found at x =  1.99999999993357E+000
Version 2: secant method:
Exact root found at x =  1.00000000000000E+000

[edit] Perl

sub f
{
my $x = shift;
 
return ($x * $x * $x - 3*$x*$x + 2*$x);
}
 
my $step = 0.001; # Smaller step values produce more accurate and precise results
my $start = -1;
my $stop = 3;
my $value = &f($start);
my $sign = $value > 0;
 
# Check for root at start
 
print "Root found at $start\n" if ( 0 == $value );
 
for( my $x = $start + $step;
$x <= $stop;
$x += $step )
{
$value = &f($x);
 
if ( 0 == $value )
{
# We hit a root
print "Root found at $x\n";
}
elsif ( ( $value > 0 ) != $sign )
{
# We passed a root
print "Root found near $x\n";
}
 
# Update our sign
$sign = ( $value > 0 );
}

[edit] Perl 6

Uses exact arithmetic.

sub f($x) { $x*$x*$x - 3*$x*$x + 2*$x }
 
my $start = -1;
my $stop = 3;
my $step = 0.001;
 
for $start, * + $step ... $stop -> $x {
state $sign = 0;
given f($x) {
my $next = .sign;
when 0.0 {
say "Root found at $x";
}
when $sign and $next != $sign {
say "Root found near $x";
}
NEXT $sign = $next;
}
}
Output:
Root found at 0
Root found at 1
Root found at 2

[edit] PicoLisp

Translation of: Clojure
(de findRoots (F Start Stop Step Eps)
(filter
'((N) (> Eps (abs (F N))))
(range Start Stop Step) ) )
 
(scl 12)
 
(mapcar round
(findRoots
'((X) (+ (*/ X X X `(* 1.0 1.0)) (*/ -3 X X 1.0) (* 2 X)))
-1.0 3.0 0.0001 0.00000001 ) )

Output:

-> ("0.000" "1.000" "2.000")

[edit] PL/I

 
f: procedure (x) returns (float (18));
declare x float (18);
return (x**3 - 3*x**2 + 2*x );
end f;
 
declare eps float, (x, y) float (18);
declare dx fixed decimal (15,13);
 
eps = 1e-12;
 
do dx = -5.03 to 5 by 0.1;
x = dx;
if sign(f(x)) ^= sign(f(dx+0.1)) then
call locate_root;
end;
 
locate_root: procedure;
declare (left, mid, right) float (18);
 
put skip list ('Looking for root in [' || x, x+0.1 || ']' );
left = x; right = dx+0.1;
PUT SKIP LIST (F(LEFT), F(RIGHT) );
if abs(f(left) ) < eps then
do; put skip list ('Found a root at x=', left); return; end;
else if abs(f(right) ) < eps then
do; put skip list ('Found a root at x=', right); return; end;
do forever;
mid = (left+right)/2;
if sign(f(mid)) = 0 then
do; put skip list ('Root found at x=', mid); return; end;
else if sign(f(left)) ^= sign(f(mid)) then
right = mid;
else
left = mid;
/* put skip list (left || right); */
if abs(right-left) < eps then
do; put skip list ('There is a root near ' ||
(left+right)/2); return;
end;
end;
end locate_root;
 

[edit] PureBasic

Translation of: C++
Procedure.d f(x.d)
ProcedureReturn x*x*x-3*x*x+2*x
EndProcedure
 
Procedure main()
OpenConsole()
Define.d StepSize= 0.001
Define.d Start=-1, stop=3
Define.d value=f(start), x=start
Define.i oldsign=Sign(value)
 
If value=0
PrintN("Root found at "+StrF(start))
EndIf
 
While x<=stop
value=f(x)
If Sign(value) <> oldsign
PrintN("Root found near "+StrF(x))
ElseIf value = 0
PrintN("Root found at "+StrF(x))
EndIf
oldsign=Sign(value)
x+StepSize
Wend
EndProcedure
 
main()

[edit] Python

Translation of: Perl
f = lambda x: x * x * x - 3 * x * x + 2 * x
 
step = 0.001 # Smaller step values produce more accurate and precise results
start = -1
stop = 3
 
sign = f(start) > 0
 
x = start
while x <= stop:
value = f(x)
 
if value == 0:
# We hit a root
print "Root found at", x
elif (value > 0) != sign:
# We passed a root
print "Root found near", x
 
# Update our sign
sign = value > 0
 
x += step

[edit] R

Translation of: Octave
f <- function(x) x^3 -3*x^2 + 2*x
 
findroots <- function(f, begin, end, tol = 1e-20, step = 0.001) {
se <- ifelse(sign(f(begin))==0, 1, sign(f(begin)))
x <- begin
while ( x <= end ) {
v <- f(x)
if ( abs(v) < tol ) {
print(sprintf("root at %f", x))
} else if ( ifelse(sign(v)==0, 1, sign(v)) != se ) {
print(sprintf("root near %f", x))
}
se <- ifelse( sign(v) == 0 , 1, sign(v))
x <- x + step
}
}
 
findroots(f, -1, 3)

[edit] Racket

 
#lang racket
 
;; Attempts to find all roots of a real-valued function f
;; in a given interval [a b] by dividing the interval into N parts
;; and using the root-finding method on each subinterval
;; which proves to contain a root.
(define (find-roots f a b
#:divisions [N 10]
#:method [method secant])
(define h (/ (- b a) N))
(for*/list ([x1 (in-range a b h)]
[x2 (in-value (+ x1 h))]
#:when (or (root? f x1)
(includes-root? f x1 x2)))
(find-root f x1 x2 #:method method)))
 
;; Finds a root of a real-valued function f
;; in a given interval [a b].
(define (find-root f a b #:method [method secant])
(cond
[(root? f a) a]
[(root? f b) b]
[else (and (includes-root? f a b) (method f a b))]))
 
;; Returns #t if x is a root of a real-valued function f
;; with absolute accuracy (tolerance).
(define (root? f x) (almost-equal? 0 (f x)))
 
;; Returns #t if interval (a b) contains a root
;; (or the odd number of roots) of a real-valued function f.
(define (includes-root? f a b) (< (* (f a) (f b)) 0))
 
;; Returns #t if a and b are equal with respect to
;; the relative accuracy (tolerance).
(define (almost-equal? a b)
(or (< (abs (+ b a)) (tolerance))
(< (abs (/ (- b a) (+ b a))) (tolerance))))
 
(define tolerance (make-parameter 5e-16))
 

Different root-finding methods

 
(define (secant f a b)
(let next ([x1 a] [y1 (f a)] [x2 b] [y2 (f b)] [n 50])
(define x3 (/ (- (* x1 y2) (* x2 y1)) (- y2 y1)))
(cond
 ; if the method din't converge within given interval
 ; switch to more robust bisection method
[(or (not (< a x3 b)) (zero? n)) (bisection f a b)]
[(almost-equal? x3 x2) x3]
[else (next x2 y2 x3 (f x3) (sub1 n))])))
 
(define (bisection f x1 x2)
(let divide ([a x1] [b x2])
(and (<= (* (f a) (f b)) 0)
(let ([c (* 0.5 (+ a b))])
(if (almost-equal? a b)
c
(or (divide a c) (divide c b)))))))
 

Examples:

 
-> (find-root (λ (x) (- 2. (* x x))) 1 2)
1.414213562373095
-> (sqrt 2)
1.4142135623730951
 
-> (define (f x) (+ (* x x x) (* -3.0 x x) (* 2.0 x)))
-> (find-roots f -3 4 #:divisions 50)
'(2.4932181969624796e-33 1.0 2.0)
 

In order to provide a comprehensive code the given solution does not optimize the number of function calls. The functional nature of Racket allows to perform the optimization without changing the main code using memoization.

Simple memoization operator

 
(define (memoized f)
(define tbl (make-hash))
(λ x
(cond [(hash-ref tbl x #f) => values]
[else (define res (apply f x))
(hash-set! tbl x res)
res])))
 

To use memoization just call

 
-> (find-roots (memoized f) -3 4 #:divisions 50)
'(2.4932181969624796e-33 1.0 2.0)
 

The profiling shows that memoization reduces the number of function calls in this example from 184 to 67 (50 calls for primary interval division and about 6 calls for each point refinement).

[edit] REXX

[edit] as-is function

/*REXX program to find the  roots  of a  specific  function.            */
parse arg bot top inc . /*allow user to specify options. */
if bot=='' | bot==',' then bot=-3 /*Not specified? Then use default*/
if top=='' | top==',' then top=+3 /* " " " " " */
if inc=='' | inc==',' then inc=.0001 /* " " " " " */
z=f(bot);  !=sign(z)
 
do j=bot to top by inc /*traipse through all the values.*/
z=f(j); $=sign(z) /*compute new value and the sign.*/
if z=0 then say 'found a root at' j/1
else if !\==$ then if !\==0 then say 'passed a root at' j/1
 !=$ /*use the new sign. */
end /*j*/
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────F function──────────────────────────*/
f: procedure; parse arg x; return x**3 - 3 * x**2 + 2 * x

output when using the defaults for input:

found a root at 0
found a root at 1
found a root at 2

[edit] optimized function

This is a slightly optimized version of the F function.

When doing thousands of evaluations, every little bit helps.

/*──────────────────────────────────F function──────────────────────────*/
f: procedure; parse arg x; x2=x*x; return x*x2 - 3*x2 + x+x

[edit] RLaB

RLaB implements a number of solvers from the GSL and the netlib that find the roots of a real or vector function of a real or vector variable. The solvers are grouped with respect whether the variable is a scalar, findroot, or a vector, findroots. Furthermore, for each group there are two types of solvers, one that does not require the derivative of the objective function (which root(s) are being sought), and one that does.

The script that finds a root of a scalar function f(x) = x^3-3\,x^2 + 2\,x of a scalar variable x using the bisection method on the interval -5 to 5 is,

 
f = function(x)
{
rval = x .^ 3 - 3 * x .^ 2 + 2 * x;
return rval;
};
 
>> findroot(f, , [-5,5])
0
 

For a detailed description of the solver and its parameters interested reader is directed to the rlabplus manual.

[edit] Ruby

Translation of: Python
def sign(x)
x <=> 0
end
 
def find_roots(f, range, step=0.001)
sign = sign(f[range.begin])
range.step(step) do |x|
value = f[x]
if value == 0
puts "Root found at #{x}"
elsif sign(value) == -sign
puts "Root found between #{x-step} and #{x}"
end
sign = sign(value)
end
end
 
f = lambda { |x| x**3 - 3*x**2 + 2*x }
find_roots(f, -1..3)
Output:
Root found at 0.0
Root found at 1.0
Root found at 2.0

Or we could use Enumerable#inject, monkey patching and block:

class Numeric
def sign
self <=> 0
end
end
 
def find_roots(range, step = 1e-3)
range.step( step ).inject( yield(range.begin).sign ) do |sign, x|
value = yield(x)
if value == 0
puts "Root found at #{x}"
elsif value.sign == -sign
puts "Root found between #{x-step} and #{x}"
end
value.sign
end
end
 
find_roots(-1..3) { |x| x**3 - 3*x**2 + 2*x }

[edit] Tcl

This simple brute force iteration marks all results, with a leading "~", as approximate. This version always reports its results as approximate because of the general limits of computation using fixed-width floating-point numbers (i.e., IEEE double-precision floats).

proc froots {lambda {start -3} {end 3} {step 0.0001}} {
set res {}
set lastsign [sgn [apply $lambda $start]]
for {set x $start} {$x <= $end} {set x [expr {$x + $step}]} {
set sign [sgn [apply $lambda $x]]
if {$sign != $lastsign} {
lappend res [format ~%.11f $x]
}
set lastsign $sign
}
return $res
}
proc sgn x {expr {($x>0) - ($x<0)}}
 
puts [froots {x {expr {$x**3 - 3*$x**2 + 2*$x}}}]

Result and timing:

/Tcl $ time ./froots.tcl
~0.00000000000 ~1.00000000000 ~2.00000000000

real    0m0.368s
user    0m0.062s
sys     0m0.030s

A more elegant solution (and faster, because you can usually make the initial search coarser) is to use brute-force iteration and then refine with Newton-Raphson, but that requires the differential of the function with respect to the search variable.

proc frootsNR {f df {start -3} {end 3} {step 0.001}} {
set res {}
set lastsign [sgn [apply $f $start]]
for {set x $start} {$x <= $end} {set x [expr {$x + $step}]} {
set sign [sgn [apply $f $x]]
if {$sign != $lastsign} {
lappend res [format ~%.15f [nr $x $f $df]]
}
set lastsign $sign
}
return $res
}
proc sgn x {expr {($x>0) - ($x<0)}}
proc nr {x1 f df} {
# Newton's method converges very rapidly indeed
for {set iters 0} {$iters < 10} {incr iters} {
set x1 [expr {
[set x0 $x1] - [apply $f $x0]/[apply $df $x0]
}]
if {$x0 == $x1} {
break
}
}
return $x1
}
 
puts [frootsNR \
{x {expr {$x**3 - 3*$x**2 + 2*$x}}} \
{x {expr {3*$x**2 - 6*$x + 2}}}]

[edit] TI-89 BASIC

Finding roots is a built-in function: zeros(x^3-3x^2+2x, x) returns {0,1,2}.

In this case, the roots are exact; inexact results are marked by decimal points.

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