Why is the polyval command giving two different answers?

Why does the polyval operator not work as expected. Is the ans variable not stored as a column vector? Why aren't the second, fifth, and sixth results equal?

>> roots([1,-8,17,2,-24])

ans =

    4.0000
    3.0000
    2.0000
   -1.0000

>> polyval([1.-8,17,2,-24],ans)

ans =

 -192.0000
  -54.0000
   -8.0000
   -2.0000

>> roots([1,-8,17,2,-24])

ans =

    4.0000
    3.0000
    2.0000
   -1.0000

>> x=ans

x =

    4.0000
    3.0000
    2.0000
   -1.0000

>> polyval([1,-8,17,2,-24],x)

ans =

   1.0e-13 *

    0.8882
    0.3197
    0.0355
    0.1421

>> polyval([1,-8,17,2,-24],[2.0000;3.0000;-1.0000;3])

ans =

     0
     0
     0
     0

Answer 1

[Alberto]

Instruction roots uses an iterative numeric method to approximate the solution in float arithmetic. What you get is an excellent approximation.

If you need the exact solution you should try a symbolic method:

g = x^4-8*x^3 + 17*x^2 +2*x -24

g =

x^4 - 8*x^3 + 17*x^2 + 2*x - 24

>> sol=solve(g==0)

sol =

2
3
4
-1

[Matt J]

You also may need a symbolic version of polyval, even when you have the exact roots:

      >> polyval([1,-8,17,2,-24]/3,[4 3 2 -1])

    ans =

       1.0e-14 *

        0.8882    0.1776    0.1776    0.1776

Answer 2

[Matt J]

Because you have a typo in your call to polyval: a period appears where a comma should be.

[Baldemy]

ok then this happens instead

>> roots([1,-8,17,2,-24])

ans =

     4.000000000000009
     2.999999999999992
     2.000000000000001
    -1.000000000000000

>> polyval([1,-8,17,2,-24],ans)

ans =

     1.0e-13 *

     0.888178419700125
     0.319744231092045
     0.035527136788005
     0.142108547152020

[Matt J]

And what don't you like about it? The result is quite close to zero, as one would expect.