Question

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

Answer 1

There are no solutions to the polynomial.

Explanation:

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.

Answer 2

`3x^2+3x+3` has `2` non-Real Complex zeros

Explanation:

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`

`color(white)()`
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.