B Confusion about The Conjugate Roots Theorem

[opus]

As a preface to this theorem stated in my text, it states that:
"If all the coefficients of a polynomial $P(x)$ are real, then $P$ is a function that transforms real numbers into other real numbers, and consequently, $P$ can be graphed in the Cartesian Coordinate Plane."

It then goes on to state The Conjugate Roots Theorem:
Let $P(x)$ be a polynomial with only real coefficients. If the complex number $a+bi$ is a zero of $P$, then so is the complex number $a-bi$. In terms of linear factors of $P$, this means that if $x-(a+bi)$ is a factor of $P$, then so is $x-(a-bi)$.

Now I understand the theorem, but what I've highlighted in bold is confusing me as it seems to be a contradiction. In the first statement, what I understand is that if we have real coefficients in a polynomial, then we'll have real numbers (and have no non-reals come up). In the theorem, it states the same given of only real coefficients, but now states that we can have non-real numbers come up. We then would not be able to graph all the solutions in the Cartesian Coordinate Plane.

Where am I thinking incorrectly here?

 

[Orodruin]

There is no contradiction here. The point is that $(x-a-ib)(x-a+ib) = x^2 - 2ax +a^2 + b^2$ so even if it is factorised it into linear factors using complex numbers, the final result will be real if $x$ is real.

Edit: Let me add that, while you can graph the polynomial, only the real roots will be zeros of the graph. For example, $y = x^2 + 1$ has no real roots so its graph never crosses the $x$-axis.

 

[opus]

There is no contradiction here. The point is that $(x-a-ib)(x-a+ib) = x^2 - 2ax +a^2 + b^2$ so even if it is factorised it into linear factors using complex numbers, the final result will be real if $x$ is real.

Edit: Let me add that, while you can graph the polynomial, only the real roots will be zeros of the graph. For example, $y = x^2 + 1$ has no real roots so its graph never crosses the $x$-axis.

What do you mean by this? How can $(x-a-ib)$ or $(x-a+ib)$ be real if they have non-real constants? I understand up to the point where you factored the polynomial into linear factors which in this case have non-real constants in them.
I do understand that we can have polynomials of degree $n$ that will have at most $n$ solutions that may be real or non-real and distinct or non-distinct. If they are non-real solutions, they will not cross the x-axis and cannot be graphed in the Cartesian Plane.

 

[Orodruin]

How can (x−a−ib)(x−a−ib)(x-a-ib) or (x−a+ib)(x−a+ib)(x-a+ib) be real if they have non-real constants?

I never said they were. I said their product is real. This follows directly from $z\bar z$ and $z+\bar z$ being real for any complex number $z$.

 

[Orodruin]

If they are non-real solutions, they will not cross the x-axis and cannot be graphed in the Cartesian Plane

What do you mean by this? Do you intend to say that the zeros will not be on the real line (true) or that you cannot graph the polynomial (false)?

 

[fresh_42]

Take $P(x)=x^2+1=(x+i)(x-i)$. It has real coefficients $1\cdot x^2 + 1$, its graph can be drawn, it doesn't cross the real line, and it has two complex and conjugate zeroes $\pm \, i\,.$

So what don't you understand in this example?

 

[slider142]

As a preface to this theorem stated in my text, it states that:
"If all the coefficients of a polynomial $P(x)$ are real, then $P$ is a function that transforms real numbers into other real numbers, and consequently, $P$ can be graphed in the Cartesian Coordinate Plane."

It then goes on to state The Conjugate Roots Theorem:
Let $P(x)$ be a polynomial with only real coefficients. If the complex number $a+bi$ is a zero of $P$, then so is the complex number $a-bi$. In terms of linear factors of $P$, this means that if $x-(a+bi)$ is a factor of $P$, then so is $x-(a-bi)$.

Now I understand the theorem, but what I've highlighted in bold is confusing me as it seems to be a contradiction. In the first statement, what I understand is that if we have real coefficients in a polynomial, then we'll have real numbers (and have no non-reals come up). In the theorem, it states the same given of only real coefficients, but now states that we can have non-real numbers come up. We then would not be able to graph all the solutions in the Cartesian Coordinate Plane.

Where am I thinking incorrectly here?

The first bolded statement means that if you use a real number as an input value, then the output value will also be a real number. It is not a prescription that bars complex numbers from being used as input values. It simply has nothing to say about what happens if you use complex numbers that are not real numbers as input values.

The second bolded statement and the associated theorem tells us what will happen if the same polynomial has a complex root (a complex input that is associated with an output of 0). Neither statement contradicts the other.

 

[opus]

I never said they were. I said their product is real. This follows directly from $z\bar z$ and $z+\bar z$ being real for any complex number $z$.

Ohh ok. The product is real. Understood.

What do you mean by this? Do you intend to say that the zeros will not be on the real line (true) or that you cannot graph the polynomial (false)?

The first one. So to my understanding, only real zeros can be represented in the Cartersian Plane as it is two intersecting number lines consisting of real numbers.

Take $P(x)=x^2+1=(x+i)(x-i)$. It has real coefficients $1\cdot x^2 + 1$, its graph can be drawn, it doesn't cross the real line, and it has two complex and conjugate zeroes $\pm \, i\,.$

So what don't you understand in this example?

I fully understand that example. I think I must have misunderstood the difference between $P(x)$ having real coefficients and the constants of the factors being non-real. In the given text, I was focusing on the statement that "If all coefficients of a polynomial are real, then $P$ is a function that transforms real numbers into real numbers", and thinking that this meant that if $P(x)$ has all real coefficients, then we can't get non-real numbers.

The first bolded statement means that if you use a real number as an input value, then the output value will also be a real number. It is not a prescription that bars complex numbers from being used as input values. It simply has nothing to say about what happens if you use complex numbers that are not real numbers as input values.

The second bolded statement and the associated theorem tells us what will happen if the same polynomial has a complex root (a complex input that is associated with an output of 0). Neither statement contradicts the other.

Yes this is exactly what was confusing me. I didn't really know how to word it correctly I guess. Sorry for the confusion guys and thanks so much for the replies.

 

[fresh_42]

You don't even need to use complex numbers. E.g. $x^2-2$ has rational coefficients, even integers, but no rational roots. Here we can draw a graph which crosses the $x-$axis twice, just at locations which aren't rational. This entire matter is subject to field theory. In this example you can add - terminus technicus is adjoint - the square root of $2$ and all its sums and rational multiples to the rationals and obtain $\mathbb{Q}[\sqrt{2}]$ which has the zeros required. But then you won't have zeros for $x^2-3$. Latest in the field of complex numbers are all roots (= zeroes) of all polynomials included. However, this doesn't mean we can write them down as expression of roots, only that they exist in there and can be numerically approximated.