Find the number of distinct complex values that $P(x) =x^6+2x^5+5x^4+4x^3+6x^2+5x+7$ if $x$ is a solution of $x^3+x+1=0$

Question

Let $n$ be the number of distinct complex values which the polynomial function $P(x) =x^6+2x^5+5x^4+4x^3+6x^2+5x+7$ takes, if $x$ is a solution of $x^3+x+1=0$. Find $n$.

I am a bit confused by this problem because what value I get for $n$ is different than what the workbook offers as a result.

If $x$ is a solution of $x^3+x+1=0$, then there is one real solution $x_1$ and two complex, conjugate solution $x_2$, $x_3$ (this can be proven by differentiating $f(x)=x^3+x+1$ once and proving $f$ crosses $OX$ once). Following this line of thought, there must be at least two complex values that $P(x)$ takes, as for $x_1$ it will be a real number and for $x_2$ or $x_3$ the value should be strictly complex (that is, with a term that contains $i$), both "outputs" being complex values (although one is real, because $\mathbb{R}\subset\mathbb{C}$), thus $n\geq 2$. Going further isn't necessary because the book says $n=1$, so there's only one complex value $P(x)$ takes using the solutions of $x^3+x+1=0$. How is that possible? Do you think by "distinct complex values" they mean purely complex numbers and not real ones? Or is perhaps my logic faulty?

Also, tangentially related, if a polynomial has real coefficients, do complex, conjugated numbers return the same value? Or can $z$ and $\bar z$ return different values?

Answer 1

You're going a bit overkill. I agree that $x^3+x+1 = 0$ has $1$ real and two complex solutions. But, they're not easy to find (you'll need the Cubic Formula). Unless you're doing a course that involves this formula (or preparing for a contest that requires it), you're not supposed to use it.

As a general method, when you know the value of a lower-degree polynomial $q(x)$ for some specific $x$, (usually $q(x)$ for these $x$s is zero, like here) and you want to find a higher degree polynomial $p(x)$, your aim should be to form an equation something like this:
$$p(x) = q(x)s(x) + r(x)$$where $r(x)$ is of the minimum possible degree (in other words, just divide $p(x)$ by $q(x)$, $s(x)$ is quotient and $r(x)$ is remainder).

Turns out, here $r(x)$ has a very less degree...

Answer 2

The following is equivalent to the long division, but the calculations may be easier to work out sometimes.

Given that $x$ is a root of $x^3 + x + 1 = 0\,$:

$$ x^3 = -x - 1 \tag{1} $$

Multiplying $(1)$ by $x\,$:

$$ x^4 = -x^2 - x \tag{2} $$

Multiplying $(2)$ by $x$ and using $(1)\,$:

$$ \begin{align} x^5 &= -x^3 - x^2 \\ &= -x^2 + x + 1 \tag{3} \end{align} $$

Multiplying $(3)$ by $x$ and using $(1)$ again:

\begin{align} x^6 &= -x^3 + x^2 + x \\ &= x^2 + 2x + 1\tag{4} \end{align}

Substituting $(1),(2),(3),(4)$ in $P(x)\,$:

$$ \require{cancel} \begin{align} P(x) &= x^6 + 2x^5 + 5x^4 + 4x^3 + 6x^2 + 5x + 7 \\ &= (\bcancel{x^2} + \cancel{2 x} + 1) + 2(-\bcancel{x^2} + \cancel{x} + 1) + 5(-\bcancel{x^2} - \cancel{x}) \\ &\;\;\;\;\;\;\;\; + 4(-\cancel{x} - 1) + \bcancel{6 x^2} + \cancel{5 x} + 7 \\ &= 6 \end{align} $$