Assuming you’re considering only a polynomial p(z) of one variable, you look at the degree, the highest power of z . The highest power of z determines the number of roots of p(z) , though some of them may be repeated roots. For example, the roots of p(z)=z2 are 0 and 0 .
Of course, that’s if p(z) is a polynomial over the field of complex numbers, and it doesn’t matter if the coefficients are real or complex.
If p is actually p(x) over the field of real numbers, with real coefficients, then the degree is merely an upper bound. If the degree is even n=2k , then there may be an even number of roots, including zero; if the degree is odd n=2k+1 , then there is an odd number of roots, with a minimum of 1 . Finding the number of roots in this case requires factorising p , and/or finding some other means of determining the number of real roots (linear algebra, graphing, testing points and putting bounds on where roots must be and then proving that within certain bounds only one root exists, et cetera).
For example, with a quadratic p(x)=ax2+bx+c , there is the discriminant Δ=b2−4ac that lets you discriminate whether or not the quadratic has solutions. If Δ>0 then there are two real roots; if Δ=0 there is one repeated real root, and if Δ<0 there are no real roots.
Answers & Comments
Answer:
Assuming you’re considering only a polynomial p(z) of one variable, you look at the degree, the highest power of z . The highest power of z determines the number of roots of p(z) , though some of them may be repeated roots. For example, the roots of p(z)=z2 are 0 and 0 .
Of course, that’s if p(z) is a polynomial over the field of complex numbers, and it doesn’t matter if the coefficients are real or complex.
If p is actually p(x) over the field of real numbers, with real coefficients, then the degree is merely an upper bound. If the degree is even n=2k , then there may be an even number of roots, including zero; if the degree is odd n=2k+1 , then there is an odd number of roots, with a minimum of 1 . Finding the number of roots in this case requires factorising p , and/or finding some other means of determining the number of real roots (linear algebra, graphing, testing points and putting bounds on where roots must be and then proving that within certain bounds only one root exists, et cetera).
For example, with a quadratic p(x)=ax2+bx+c , there is the discriminant Δ=b2−4ac that lets you discriminate whether or not the quadratic has solutions. If Δ>0 then there are two real roots; if Δ=0 there is one repeated real root, and if Δ<0 there are no real roots.
Step-by-step explanation: