How many roots does the polynomial #3x^2+3x+3# have?
2 Answers
Answer:
There are no solutions to the polynomial.
Explanation:
First of all, you can cancel out the constant
Answer:
Explanation:
Fundamental Theorem of Algebra
The Fundamental Theorem of Algebra (FTOA) tells us that a (single variable) polynomial of degree
A straightforward corollary of the FTOA, often stated with it, is that any polynomial of degree
In our example, the given polynomial is of degree
So the FTOA gives us a quick way of answering such a question, but is itself not easy to prove.
Discriminant
Given a quadratic in the form
#Delta = b^24ac#
Then:

If
#Delta > 0# then#f(x)# has two distinct Real zeros. 
If
#Delta = 0# then#f(x)# has one repeated Real zero (of multiplicity#2# ). 
If
#Delta < 0# then#f(x)# has two distinct nonReal Complex zeros, which are Complex conjugates of one another.
In our example, we could just plug in
#Delta = 3^24(3)(3) = 936 = 27#
So our quadratic has no Real zeros. It has a Complex conjugate pair of nonReal zeros.
Quadratic formula
The discriminant
#x = (b + sqrt(b^24ac))/(2a)#
#color(white)(x) = (b+sqrt(Delta))/(2a)#
#color(white)(x) = (3+sqrt(27))/(2*3)#
#color(white)(x) = (3+3sqrt(3)i)/6#
#color(white)(x) = 1/2+sqrt(3)/2i#
Footnote
If you multiply the given quadratic by
So the zeros we found must be cube roots of
In fact they are the two nonReal Complex cube roots of
It is very useful when finding the zeros of cubic polynomials.