How many roots does the polynomial #3x^2+3x+3# have?

First of all, you can cancel out the constant #3# which will get you #x^2+x+1#. After this, to check for how many roots this has, we can use the discriminant #b^2-4ac#. If this value is #0#, then the polynomial has #1# solution. If the value is positive, the polynomial has #2# solutions. If the value is negative, than the polynomial has no solutions. In this case, we have #1^2 - 4(1)(1)=1-4=-3# Because our result is negative, there are no solutions to this polynomial.

Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra (FTOA) tells us that a (single variable) polynomial of degree #n > 0# with Complex (possibly Real) coefficients has a Complex (possibly Real) zero.

A straightforward corollary of the FTOA, often stated with it, is that any polynomial of degree #n > 0# has exactly #n# zeros counting multiplicity.

In our example, the given polynomial is of degree #2# so has exactly #2# zeros counting multiplicity.

So the FTOA gives us a quick way of answering such a question, but is itself not easy to prove.

#color(white)()#
Discriminant

Given a quadratic in the form #f(x) = ax^2+bx+c# with #a, b, c# Real coefficients, we can find out about the nature of its zeros by looking at its discriminant #Delta#, given by the formula:

#Delta = b^2-4ac#

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 non-Real Complex zeros, which are Complex conjugates of one another.

In our example, we could just plug in #a=b=c=3# to find:

#Delta = 3^2-4(3)(3) = 9-36 = -27#

So our quadratic has no Real zeros. It has a Complex conjugate pair of non-Real zeros.

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Quadratic formula

The discriminant #Delta# given above occurs in the quadratic formula, which immediately gives us the zeros:

#x = (-b +- sqrt(b^2-4ac))/(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#

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Footnote

If you multiply the given quadratic by #(x-1)/3# then you get #x^3-1#

So the zeros we found must be cube roots of #1#.

In fact they are the two non-Real Complex cube roots of #1#, sometimes denoted #omega = -1/2+sqrt(3)i = cos((2pi)/3)+ i sin((2pi)/3)# and its conjugate #bar(omega) = -1/2-sqrt(3)i#

#omega# is called the primitive Complex cube root of #1#

It is very useful when finding the zeros of cubic polynomials.