Expert: Josh - 3/7/2005

Question
HOw many possible real roots are there for x^5 - 4x^4 + 3x^3 - 2x + 2.
My teacher said that there are either 5,4,3,2,1, or 0 possible roots then he said something like that there can't be 4, 2 or 0 real roots because of something like if there are 4 real roots, it needs 1 imaginary and we know that you can't have 1 imaginary because they come in conjugates.
I don't understand why if you would have 4 real roots, you would need 1 imaginary.Can you please explain?Thank you so much.

Answer
This is a consequence of the fundamental theorem of algebra (F.T.O.A). It states that given a polynomial of degree N, with real-valued coefficients, there exists precisely N complex-valued solutions.

Effectively, it can be factorized into (z-z1)(z-z2)(z-z3)...(z-z_N), where z1,z2,z3,...,z_N are generally complex numbers.

Special case: If the argument (i.e., angle subtended by the vector) of a complex root is 0 or 180 degrees (as measured in the counter-clockwise direction from the positive end of the real-axis), then, its imaginary component is zero. It will be something like a_k + i* b_k, with b_k=0. Under this condition, we have a real root.

Reflect on the lessons you have learnt in solving quadratic equations. In particular, the solutions have the form z=[-b+sqrt(b^2-4ac)]/(2a),z=[-b-sqrt(b^2-4ac)]/(2a). Whether the roots of the quadratic is real or not, is determined solely by the sign of the argument inside the square root. If b^2-4ac<0, we can always write the solutions as z=[-b+i*d]/2a,[-b-i*d]/2a, where d=sqrt(4ac-b^2). Here, I have used the fact that i*i=sqrt(-1)*sqrt(-1)=-1. So, for example, if you have sqrt(-4), sqrt(-4)=sqrt(-1*4)=sqrt(-1)*sqrt(4)=2i.

Implications: For a quadratic equation, it has either
(i) two distinct real-valued solutions (of the form z=[-b+sqrt(.)]/(2a) and [-b-sqrt(.)]/(2a)) OR
(ii) a root of multiplicity two (in this case, it must be real-valued as evident from the quadratic formula, z=-b/(2a),-b/(2a), i.e., same root appears twice) OR
(iii) no real solution (i.e., both roots are complex-valued, z=[-b+i*d]/2a,[-b-i*d]/2a, where i=sqrt(-1)).

Due to the F.T.O.A., we can build higher degree polynomials using only linear and quadratic factors (with real-valued coefficients). This is equivalent to saying that all polynomials of degree N, with real-valued coefficients, can be factorized into
(i) N linear factors over the complex field, (i.e., allowing the roots to be either real or complex-valued) OR
(ii) If N=2M+1 is odd, we can factorized it into 2M quadratic factors (each with real-valued coefficient) plus a linear factor.

In case (ii), some of the 2M quadratic factors may have real solutions and therefore can be decomposed into linear factors. In general, the roots for each quadratic polynomial may be written as (z-z1)(z-z1*), where if z1=a+ib, then, z1*=a-ib.

Because (z-z1)(z-z1*) is guaranteed to expand to a real-valued polynomial, if we multiply this by (z-z3), where z3 is complex-valued, there is no way that the resulting cubic polynomial could have all real-valued coefficients.

e.g., suppose that z1=1+sqrt(2),z1*=1-sqrt(2),
(z-z1)(z-z1*)=z^2+2z+5
if z3 is a complex root, (z+a+ib)(z^2+2z+5) must contain some imaginary term. Since we work with real-valued polynomial with degree N=2M+1, we cannot allow z3 to be complex-valued. z3 must be real
Conclusion - The cubic equation must have at least one real-solution.
We cannot have 0 or 2 real solutions.
We have either 1 or 3 real solutions.

We can build on this argument by introducing more quadratic factors. A polynomial of degree N=5 has the form
(z-z1)(z-z1*)(z-z3)(z-z3*)(z-z5).
The F.T.O.A. guarantees that (z-z1)(z-z1*)(z-z3)(z-z3*) is a polynomial with real-coefficients. In general, if z1=a1+i*b1 and it conjugate root, z1*=a1-i*b1, (z-z1)(z-z1*) always expands to z^2+2*Real(z1)*z+|z1|^2.
Both Real(z1)=Real(a1+i*b1)=a1 and the modulus |z1|=z1.z1*=sqrt(a1^2+b1^2) are real-valued.

Send me a follow-up and tell me if there's anything that you don't understand.

Cheers.