Polynomial question (2 Viewers)

Skeptyks · Jan 30, 2012

Edit: New question. It says by using 2 and 1 + i as roots, construct a polynomial with the lowest degree with rational coefficients. If it has complex coefficients, then the highest degree is 2 but why is the highest degree 3 when you are dealing with rationals?

Solving z^4 - z^3 + 6z^2 - z + 15 = 0 for z given z = 1-2i is a root.
So, z = 1+21 because of real coefficients, b=/= 0 is also a root.
Letting the other roots be (in this case, no latex) A and B.
Sum of roots: A + B = -1
Product of Roots: AB = 3

Now, the answers proceed to say, 'therefore by constructing a new equation with A and B as roots, we get: z^2 + z + 3 = 0 and they solve it.
I would normally just solve simultaneously but there might be a problem with that and this step looks slightly easier. How do they "construct" this new equation? I partially remember something about x^2 + (A+B)x + (AB) or something similar to that.

Any help is appreciated, sorry for the weak question

P

SpiralFlex · Jan 30, 2012

Skeptyks said:
Solving z^4 - z^3 + 6z^2 - z + 15 = 0 given z = 1-2i is a root.
So, z = 1+21 because of real coefficients, b=/= 0 is also a root.
Letting the other roots be (in this case, no latex) A and B.
Sum of roots: A + B = -1
Product of Roots: AB = 3

Now, the answers proceed to say, 'therefore by constructing a new equation with A and B as roots, we get: z^2 + z + 3 = 0 and they solve it.
I would normally just solve simultaneously but there might be a problem with that and this step looks slightly easier. How do they "construct" this new equation? I partially remember something about x^2 + (A+B)x + (AB) or something similar to that.

Any help is appreciated, sorry for the weak question P

Correct. This is an excellent method to solve the equations.

By constructing another equation you can find the roots with ease.

If $\alpha, \beta$ are the roots.

Then a quadratic equation will take the form of:

$y=(x-\alpha)(x-\beta)$

Assuming it's monic.

If I expand this, I would get,

$y=x^2-(\alpha+\beta)x+\alpha \beta$

$y=x^2-$(Sum of the roots)$x+$(Product of the roots)$$

Now in your case (monic):

$y=x^2-(-1)x+3$

$y=x^2+x+3$

Now using the quadratic formula,

$x=\frac{-1\pm 11i}{2}$

$\therefore \alpha, \beta =\frac{-1\pm 11i}{2}$

So the roots are $\frac{-1\pm 11i}{2}, 1\pm 2i$

Sometimes you would get some complicated simultaneous equation and constructing the new equation and solving it via the quadratic formula will help you immensely!

Skeptyks · Jan 30, 2012

Thanks heaps for your crazily fast reply (how can you do that while typing with latex). I knew that I was missing something from that equation I wrote up there, it was the negative there but at least I understand now

I can't believe I forgot something fundamental like this -.-

SpiralFlex · Jan 30, 2012

I am cute.

Skeptyks · Jan 30, 2012

One more question please

Edited above.

SpiralFlex · Jan 30, 2012

With rationals you are dealing with real numbers and hence conjugate root theorem is applicable so you get three roots. Sorry on iPhone can't type equations

Carrotsticks · Jan 30, 2012

I have a rather different method:

$\\ $With some trivial calculations using sum/product of roots, the quadratic with roots $ 1+2i $ and $ 1-2i $ is $ z^{2}-2z+5 \\\\ $We observe that: $ \\ \begin{align*} P(z)&=z^{4} - z^{3} +6z^{2}-z+15 \\ &= P_1(z) \left ( z^{2}-2z+5 \right ) \end{align*} \\\\ $With some simple inspection, we see that $ P_1(z) = z^{2}+z+3 \\\\ $Using the quadratic formula on $ P_1(z) $ obtains the other two solutions.$$

Skeptyks · Jan 30, 2012

Oh, of course. Thanks for that, 1 last question, I promise. In the next part, the solutions (for the rational bit), start off by saying if 1 + root 3 is a root, than so is 1 - root 3. Surely they aren't using the conjugate root theorem right? Since it isn't even a complex number.

Skeptyks · Jan 30, 2012

Carrotsticks said:
I have a rather different method:

$\\ $With some trivial calculations using sum/product of roots, the quadratic with roots $ 1+2i $ and $ 1-2i $ is $ z^{2}-2z+5 \\\\ $We observe that: $ \\ \begin{align*} P(z)&=z^{4} - z^{3} +6z^{2}-z+15 \\ &= P_1(z) \left ( z^{2}-2z+5 \right ) \end{align*} \\\\ $With some simple inspection, we see that $ P_1(z) = z^{2}+z+3 \\\\ $Using the quadratic formula on $ P_1(z) $ obtains the other two solutions.$$

Thanks, that's a pretty nice method. One thing though, wouldn't it be a bit confusing once you get some more complex numbers. After that, the long division bit will get a bit weird. Correct me if I'm wrong though

SpiralFlex · Jan 30, 2012

First a real number belongs to the complex field. So they are complex numbers. However you are right in saying its not the conjugaTe root theorem. They are conjugate surds. Quadratics have conjugate surds as their roots. So hard typing on iPhone

Carrotsticks · Jan 30, 2012

Skeptyks said:
Oh, of course. Thanks for that, 1 last question, I promise. In the next part, the solutions (for the rational bit), start off by saying if 1 + root 3 is a root, than so is 1 - root 3. Surely they aren't using the conjugate root theorem right? Since it isn't even a complex number.

No need for 'Last question, I promise'. Post as many questions as you like, and we will answer them, provided that you've shown that you've put some thought into it and genuinely cannot continue.

No, it is not the conjugate root theorem.

SpiralFlex · Jan 30, 2012

Carrotsticks said:
I have a rather different method:

$\\ $With some trivial calculations using sum/product of roots, the quadratic with roots $ 1+2i $ and $ 1-2i $ is $ z^{2}-2z+5 \\\\ $We observe that: $ \\ \begin{align*} P(z)&=z^{4} - z^{3} +6z^{2}-z+15 \\ &= P_1(z) \left ( z^{2}-2z+5 \right ) \end{align*} \\\\ $With some simple inspection, we see that $ P_1(z) = z^{2}+z+3 \\\\ $Using the quadratic formula on $ P_1(z) $ obtains the other two solutions.$$

Lol that's basically doing equating coefficients method in your head

Skeptyks · Jan 30, 2012

SpiralFlex said:
First a real number belongs to the complex field. So they are complex numbers. However you are right in saying its not the conjugaTe root theorem. They are conjugate surds. Quadratics have conjugate surds as their roots. So hard typing on iPhone

Carrotsticks said:
No need for 'Last question, I promise'. Post as many questions as you like, and we will answer them, provided that you've shown that you've put some thought into it and genuinely cannot continue.

No, it is not the conjugate root theorem.

Thankyou :] I was going to get some sleep anyway. I genuinely didn't know that quadratics have conjugate surds as their roots.

Carrotsticks · Jan 30, 2012

SpiralFlex said:
Lol that's basically doing equating coefficients method in your head

Yes, I am equating coefficients in my head but by inspection, you can perform any complete polynomial long division in a matter of seconds.

Skeptyks · Jan 30, 2012

So the question is write down an equation of the lowest possible degree with i) complex coefficients, II) rational coefficients and having the following among its roots.
It gives you <a href="http://www.codecogs.com/eqnedit.php?latex=\sqrt{3} @plus; 1 \and \2-i" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\sqrt{3} + 1 \and \2-i" title="\sqrt{3} + 1 \and \2-i" /></a>.

I can do the complex coefficients without a problem but I am just a bit confused with the second part. When I attempt it, I first get <a href="http://www.codecogs.com/eqnedit.php?latex=\sqrt{3} - 1 \and \2@plus;i" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\sqrt{3} - 1 \and \2+i" title="\sqrt{3} - 1 \and \2+i" /></a> as well as the first two roots they gave you, because of conjugate surds/roots. It also has a lowest degree of 4 because of this. So first, I perform a sum of roots which leads to <a href="http://www.codecogs.com/eqnedit.php?latex=\sqrt{3} - 1 @plus;\sqrt{3}-1 \and \2@plus;i @plus;2 - i" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\sqrt{3} - 1 +\sqrt{3}-1 \and \2+i +2 - i" title="\sqrt{3} - 1 +\sqrt{3}-1 \and \2+i +2 - i" /></a>. This gives <a href="http://www.codecogs.com/eqnedit.php?latex=3\sqrt{2} @plus; 4" target="_blank"><img src="http://latex.codecogs.com/gif.latex?3\sqrt{2} + 4" title="3\sqrt{2} + 4" /></a>.
The solutions, however, makes the roots <a href="http://www.codecogs.com/eqnedit.php?latex=1@plus;\sqrt{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?1+\sqrt{3}" title="1+\sqrt{3}" /></a> and <a href="http://www.codecogs.com/eqnedit.php?latex=1-\sqrt{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?1-\sqrt{3}" title="1-\sqrt{3}" /></a> giving a nice answer of 6 when performing a sum of roots. Is it simply because conjugate surds are placed in this format ( a + b^1/2) or is it because they just can do this for convenience?

Oh god, I was searching up how to use latex and it said add these tex things... I am so fail one sec while I fix.

Carrotsticks · Jan 30, 2012

^^^ It's because the end tag is [/tex] not [./tex]

Skeptyks · Jan 31, 2012

Oh, I figured out that you can just copy + paste the link they provide you with. Oh well

Carrotsticks · Jan 31, 2012

It says rational coefficients, so we use the Conjugate root theorem. This works for only complex roots, not real roots. This means $\sqrt{3}-1$ does not have to be a root. However, 2-i and 2+i MUST be roots (since rational coefficients). Since we are looking for the lowest possible degree, adding it as a root would contradict our goal.

Skeptyks · Jan 31, 2012

The answer does go and add 1 + 3^1/2 and 1 - 3^1/2. Where did they get the 1 - 3^1/2 from then?
They say, since 1 + 3^1/2 is a root, then so is 1 - 3^1/2.

Carrotsticks · Jan 31, 2012

Oh, I forgot about the rational part. Since its rational, we need to have the conjugate even for the real solution (in order to thus make a rational factor, resulting in a rational polynomial).

Was reading it as 'real' for some reason.

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