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

Dec 9, 2016

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 ${b}^{2} - 4 a c$. 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 ${1}^{2} - 4 \left(1\right) \left(1\right) = 1 - 4 = - 3$ Because our result is negative, there are no solutions to this polynomial.

Dec 10, 2016

$3 {x}^{2} + 3 x + 3$ 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.

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Discriminant

Given a quadratic in the form $f \left(x\right) = a {x}^{2} + b x + c$ 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:

$\Delta = {b}^{2} - 4 a c$

Then:

• If $\Delta > 0$ then $f \left(x\right)$ has two distinct Real zeros.

• If $\Delta = 0$ then $f \left(x\right)$ has one repeated Real zero (of multiplicity $2$).

• If $\Delta < 0$ then $f \left(x\right)$ 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:

$\Delta = {3}^{2} - 4 \left(3\right) \left(3\right) = 9 - 36 = - 27$

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

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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- 3 \pm \sqrt{- 27}}{2 \cdot 3}$

$\textcolor{w h i t e}{x} = \frac{- 3 \pm 3 \sqrt{3} i}{6}$

$\textcolor{w h i t e}{x} = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

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Footnote

If you multiply the given quadratic by $\frac{x - 1}{3}$ 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 $\omega = - \frac{1}{2} + \sqrt{3} i = \cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)$ and its conjugate $\overline{\omega} = - \frac{1}{2} - \sqrt{3} i$

$\omega$ is called the primitive Complex cube root of $1$

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