Real Roots of a Polynomial

I have used "solve" to factor a fourth order polynomial. It has four roots with three complex numbers. I used:

xf = solve(x^4+7*x^3-8*x^2+5*x+2,x)
if (isreal(xf)==1)
  ...
end;

to try to pull out the real roots but it did not work.

Is there a better way?

[Matt Kindig]

Are you sure there are real roots to the polynomial? It is not a given that there are.

[Azzi Abdelmalek]

Polynomial with real coefficient can not have an odd number of complex roots,

Answer 1

[Roger Stafford]

Use 'roots' to find the roots of polynomials.

 r = roots([1,7,-8,5,1]); % Get all the roots
 r = r(imag(r)==0); % Save only the real roots

The 'isreal' function is true only if All elements of a vector are real, so it isn't appropriate for sorting out the real roots.

A polynomial with all real coefficients such as yours cannot have an odd number of complex roots. They must occur in conjugate pairs. As you see, in your particular polynomial there are just two complex roots, which are conjugates of one another.

[Philosophaie]

I get:

acp1 =
      [ empty sym ]

 There is a real root to the polynomial in question.
It works my mistake.

Is there any way of getting Matlab to complete the addition,subtraction, mult and divide?

Answer 2

[Azzi Abdelmalek]

%P=x^4+7*x^3-8*x^2+5*x+2

p=[1 7 -8 5 2]
result=roots(p)

[Azzi Abdelmalek]

If all your roots are real,

syms x
factor(x^4+7*x^3-8*x^2+5*x+2)

should work

[Roger Stafford]

Only two of the roots are real. The other two are a complex conjugate pair. However the polynomial can still be factored using these complex values.