Proving (in C) that a polynomial with n=3 has 3 real roots if...

So I have to show 2 things.

i) If:

has 2 real roots (say x1 and x2), then show (in C) that x3 is also a real root.

I know from intuition this is obvious , since a cubed function that hits the x axis twice MUST go back in the other direction, but Im having troubling proving it. I have tried a couple things with no success.

Thoughts, I know that P(x)=(x-x1)(x-x2)(x-x3) but that leads me no where, and I also know that:

-a=x1+x2+x3

b=x1x2+x1x3+x2x3

-c=x1x2x3

ii) is giving me more problems. it asks to give a formula for x3 in terms of x1 and x2.

Any help or guidance is much appreciates.

Thanks

Re: Proving (in C) that a polynomial with n=3 has 3 real roots if...

what you need:

complex roots of a real polynomial occur in conjugate pairs.

proof:

in C, we have (z+w)* = z*+w* and (zw)* = z*w*.

suppose p(x) is a real polynomial, say p(x) = a_{0} + a_{1}x +....+ a_{n}x^{n}.

suppose further that z is a complex number with p(z) = 0. we want to show that p(z*) = 0, as well. remember, for a real number a, a* = a (there's no imaginary part to change sign).

now p(z*) = a_{0} + a_{1}z* +....+ a_{n}(z*)^{n}

= (a_{0})* + (a_{1})*z* +...+ (a_{n})*(z^{n})* (since the a_{j} are real, and (z*)^{n} = (z*)(z*)...(z*) = [(z)(z)...(z)]*)

= (a_{0})* + (a_{1}z)* +...+ (a_{n}z^{n})* (since the conjugate of the product is the product of the conjugates)

= (a_{0} + a_{1}z +...+ a_{n}z^{n})* (since the conjugate of the sum is the sum of the conjugates)

= (p(z))* = 0* = 0 (since 0 is real).

now what does this mean for a cubic?

if one root is complex (but not real), then TWO roots are complex.

but if we already have 2 real roots, then the third root must be its own conjugate (since we have at MOST 3 roots, and thus don't have room for a conjugate-pair), and so it must be real.

that is the only possible configurations are:

complex, complex conjugate of the first root, real root.

real, real, real.

(we either have one conjugate pair, or none).

Re: Proving (in C) that a polynomial with n=3 has 3 real roots if...

So let me get this straight, we (you) proved that, if some p(z) in C, is a root, then its conjugate z* must also be a root since z*=0*=0 since theres no imaginary part to change sign. So since there are 2 real roots already, there isnt any "room" for a conjugate pair, therefor the last root must be real.

Re: Proving (in C) that a polynomial with n=3 has 3 real roots if...

for part ii)

I might be on to something.

Quote:

P(x1)=(x1)^3-(x1+x2+x3)(x^2)+x1(x1x2+..+x2x3)-x1x2x3=0

Im not sure if this is right, but after a lot of cancelling out I end up with

Quote:

-2(x1^2)x2=2(x1^2)x3

So x3=-x2?

This cant be right and im sure i divided by 0 somewhere?

Re: Proving (in C) that a polynomial with n=3 has 3 real roots if...

Quote:

I know from intuition this is obvious , since a cubed function that hits the x axis twice MUST go back in the other direction, but Im having troubling proving it. I have tried a couple things with no success.

If something is obvious, try to prove the way your intuition says. Play around with the end behavior limits of cubic polynomials and the intermediate value theorem and see what you get.

Re: Proving (in C) that a polynomial with n=3 has 3 real roots if...

Quote:

Originally Posted by

**calculuskid1** So let me get this straight, we (you) proved that, if some p(z) in C, is a root, then its conjugate z* must also be a root since z*=0*=0 since theres no imaginary part to change sign. So since there are 2 real roots already, there isnt any "room" for a conjugate pair, therefor the last root must be real.

z* isn't 0, p(z*) is 0.

a cubic has 3 roots (at most, for example (x-1)^{3} just has one root, 1. exactly 3 roots if you count "repeated roots").

the two real roots already ARE "conjugate-pairs" (with themselves). if a root r is real, then r+0i = r-0i = r (that is r* = r). we have one root left, and whatever it is, (say z, for now), z* must also be a root. since we only have one root-slot left, z and z* must both be filling it, that is: z = z*.

so let's look at this: suppose z = a+bi. then z* = a-bi.

if z = z*, then a+bi = a-bi. subtracting a from both sides, bi = -bi.

multiplying by 1/i (or -i, they're the same), we get b = -b, so 2b = 0, so b = (2b)/2 = 0/2 = 0. so z = a+0i = a, that is: z is real.

***********

a "calculus-style" proof IS possible, but there are pit-falls:

it could be that even though we have two real roots (x_{1}, x_{2}), p'(x_{2}) is a local minimum (for a positive leading coefficient), so we can't apply the intermediate value theorem in this case, because p(x) never crosses the x-axis again. so it needs to be shown that p(x) and p'(x) share a root if and only if p(x) has a repeated root.

(the if part is easy, the only if part is a bit harder).

if the 3 roots are distinct, then p'(x) isn't 0 at x_{2}. so it must be > 0 or < 0. by rolle's theorem, p'(x) is 0 somewhere in between x_{1} and x_{2}. let's call this point c.

now let's think about p'(x) for a second. it's a quadratic, and c is a root. suppose it were the SECOND root of p'(x). then there must be some b < c, with p'(b) = 0.

so p(x) is increasing on (-∞,b) and decreasing on (b,c), and then increasing again on (c,∞). since p(x) crosses the x-axis only once on (-∞,c), p(c) > 0. but if p(x) is increasing on (c,∞), it never crosses the x-axis again, and we know it does (at x_{2}).

so c cannot be the second root of p'(x), it must be the first. so p(x) is decreasing on (c,d) where d is the second root of p'(x) (and there must BE a second root, because if not, p(x) is increasing on (c,∞), and that means p(x) never crosses the x-axis again).

if d < x_{2}, then p(x) never crosses the x-axis for the second time. thus d > x_{2}, so p'(x) is negative on (c,d),

so p'(x_{2}) < 0, and on (x_{2},d) p(x) is decreasing. so, for example at y = (x_{2} + (d-x_{2})/2), we have p(y) < 0,

and for some LARGE integer N, we have p(N) > 0, so p(x) = 0 at some x in (y,N) by the intermediate-value theorem.

this is only "half-rigorous" but i hope it illustrates some of the intricacies we have to look out for when using calculus to prove things about polynomials.