How does one interpret "the product of three roots"

Question

In a question in my test, it asked me about "the product of three roots of $x^3-3x+2$.

However, $x^3-3x+2$ does not have three roots, it has $2$ roots, $1$ and $-2$. So how should I interpret this question?

Answer 1

Factoring, we have $x^3-3x+2=(x-1)^2(x+2)$. So it has three roots: $1, 1, -2$. That is, $1$ is a root with multiplicity $2$.

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This may seem odd at first, but it turns out (for a variety of reasons) to be the "right" way to count the roots of a polynomial.

Answer 2

It is generally known that for a polynomial of degree $n $ given by $$P = a_0 x^n + a_1 x^{n-1 } + a_2 x^{n-2} +\cdots + a_n$$ , the product of all its $n $ roots is given by $$x_1x _2\cdots x_n = (-1)^n \frac {a_n}{a_0}$$ It is called Vieta's formula.

Applying this concept to the given polynomial $P = x^3 +0x^2-3x+2$, we get $$x_1x_2x_3 = -2$$ Observe that it is a cubic polynomial so it will definitely have three roots, real or complex, repeated or unrepeated. Observe that factoring our polynomial results in a repeated root, that is, $1$. Hope it helps.

Answer 3

As your question was already addressed, I'm going to mention (in my eyes) a cool way to answer this quickly using some higher-level math called Linear Algebra.

In linear algebra, we look at things called Matrices in detail. What these are aren't super important, but one of the things we associate with a matrix is called its characteristic polynomial. The zeros of this (typically denoted $\lambda_i$ for various values of $i$) are important, and are called the eigenvalues of the matrix. There are two quantities associated with the eigenvalues (or roots) that are especially important, called the Trace: $$\text{Trace}(A)= \lambda_1+\lambda_2+\dots+\lambda_n$$ and the Determinant: $$\det(A) = \lambda_1\lambda_2\dots\lambda_n$$ (Here $A$ is one of those matrix things I mentioned, again not too important).

As I said before, $\lambda$'s are just the roots of some polynomial, and the determinant is the product of all of them! We can use this to our advantage, because it's known that all characteristic polynomials are of the form: $$f(x) = x^n-\text{Trace}(A)x^{n-1}+\dots+(-1)^n\det(A)$$

Notice that your polynomial is: $$f(x) = x^3-3x-(-1)^32$$ Is this the characteristic polynomial of some matrix? It turns out yes! So, for that matrix $A$, we have that $\det(A) = \lambda_1\lambda_2\lambda_3 = -2$, so the product of the roots is $-2$.