# Hey where do i begin with this complex polynomial question?

1. Apr 8, 2014

### ivan_x3000

1. The problem statement, all variables and given/known data
Let f(x), g(x), and h(x) be polynomials in x with real coefficients. Show that if (f(x))^2 −x(g(x))^2 =x(h(x))^2,

then f(x) = g(x) = h(x) = 0. Find an example where this is not the case when we use polynomials with complex coefficients.

i have no idea where to start here

2. Relevant equations
long division
euclid's algorithm
alpha, beta, gama roots
complex form

3. The attempt at a solution
I have no idea where to start here

2. Apr 8, 2014

### Ray Vickson

For real polynomials, $g(x)^2$ and $h(x)^2$ are polynomials of even degree and with leading coefficients > 0, so this is true of $g(x)^2 + h(x)^2$ as well. The polynomial $x g(x)^2 + x h(x)^2 = x [(g(x)^2 + (h(x)^2]$ has odd degree with a positive leading coefficient, so cannot be equal to the even-degree polynomial $f(x)^2$.

See if you can spot the place or places where the assumption of real coefficients was used.

3. Apr 8, 2014

### jbunniii

Maybe start by comparing the degrees of $(f(x))^2$, $x(g(x))^2$, and $x(h(x))^2$.

 Ray beat me to it.

4. Apr 8, 2014

### Ray Vickson

Your response is better: it does not give away as much as mine. I overdid it there; sorry!

5. Apr 8, 2014

### ivan_x3000

wait if i said the polynomial was x

x^2 -x^3 = x^3
they have the same degree but they are not equal, really confused by this whole complex number thing

what does f(x)=g(x)=h(x)=0 entail?

So...
f(x)^2 = x(g(x)^2+h(x)^2)
disproves the equation because the degree of lhs has to be even and the degree of rhs has to be odd

So the only way the assumption would be true is if it were all equal to zero oh ok

6. Apr 8, 2014

### jbunniii

Yes, that's the point. The goal is to show that with real coefficients, you only get equality if $f$, $g$, and $h$ are all identically zero. Your example is consistent with this.

Yes, that's right. But be sure to write out the details, because this is false if the coefficients are allowed to be complex: in that case, you can get equality even if the polynomials are not all zero.

7. Apr 8, 2014

### ivan_x3000

But if i say... they all equal to (1+i)

(1+i)^2 + x(1+i)^2 = x(1+i)^2

how do i deal with the x's

8. Apr 8, 2014

### Ray Vickson

Polynomials having the same degree may be (identically) equal or unequal; your illustration above is for the unequal case.

Polynomials that are equal must have the same degree; in other words, if the degrees are different the polynomials cannot be (identically) equal.

9. Apr 8, 2014

### jbunniii

No, that example doesn't work. Try choosing $g(x)$ and $h(x)$ so that the leading coefficients of $x(g(x))^2$ and $x(h(x))^2$ cancel each other out (one is the negative of the other).

10. Apr 8, 2014

### ivan_x3000

how about x=i, f(x)=1+i, g(x)=1+i, h(x)=1-i

(1+i)^2 + x(1+i)^2 = = x(1-i)^2
(1+i)^2 = (1+i+1-i)(1^2-i+i^2)
2i = x(2i) which if x=1 then it's true?

11. Apr 8, 2014

### HallsofIvy

Staff Emeritus
Misunderstood the question!

12. Apr 8, 2014

### ivan_x3000

Wait how come? I'm trying to do the part where i have to show how the assumption is through for polynomials in the complex plane

13. Apr 8, 2014

### jbunniii

If you are trying to show equality of polynomials, you don't pick a value for $x$. The equality has to be true for all values of $x$.