MHB Polynomial that includes complex numbers

[PurpleDude]

Hello everyone, I'm new to this forum. I have this Linear Algebra question that I have no clue how to solve. Any help would be much appreciated. :)

The question goes as follows:

The polynomial p(x) = x³ + kx + (3 - 2i)

where k is an unknown complex number. It is given to you that if p(x) is divided by
2x - (4 - i), then the remainder is 5 + i. Find the value of k.

My question is, how would I even start this kind of problem? Again, thanks for the help!

 

[mente oscura]

Hello everyone, I'm new to this forum. I have this Linear Algebra question that I have no clue how to solve. Any help would be much appreciated. :)

The question goes as follows:

The polynomial p(x) = x³ + kx + (3 - 2i)

where k is an unknown complex number. It is given to you that if p(x) is divided by
2x - (4 - i), then the remainder is 5 + i. Find the value of k.

My question is, how would I even start this kind of problem? Again, thanks for the help!

Hello.

I only see divide.

\dfrac{x^3+kx+(3-2i)}{2x-(4-i)}=

=\dfrac{x^2}{2}+\dfrac{x}{4}(4-i)+\dfrac{1}{2}[\dfrac{(4-i)^2}{4}+k]+Rest

Rest=\dfrac{1}{2}[\dfrac{(4-i)^2}{4}+k](4-i)+(3-2i)

Therefore:

Rest=\dfrac{1}{2}[\dfrac{(4-i)^2}{4}+k](4-i)+(3-2i)=5+i

\dfrac{(4-i)^3}{8}+\dfrac{k(4-i)}{2}+3-2i=5+i

(4-i)^3+4k(4-i)=16+24i

64-48i+12i^2-i^3+4k(4-i)=16+24i

64-48i-12+i+4k(4-i)=16+24i

4k(4-i)=-36+71i

Let \ k=a+bi

4(a+bi)(4-i)=-36+71i

16a-4ai+16bi-4bi^2=-36+71i

16a-4ai+16bi+4b=-36+71i

Therefore:

16a+4b=-36

-4ai+16bi=71i \rightarrow{}-4a+16b=71

a=-\dfrac{215}{68}

b=\dfrac{62}{17}

Solution:

k=- \dfrac{215}{68}+\dfrac{62}{17}i

I hope to not have me wrong calculations.

Regards.

 

[MarkFL]

Hello everyone, I'm new to this forum. I have this Linear Algebra question that I have no clue how to solve. Any help would be much appreciated. :)

The question goes as follows:

The polynomial p(x) = x³ + kx + (3 - 2i)

where k is an unknown complex number. It is given to you that if p(x) is divided by
2x - (4 - i), then the remainder is 5 + i. Find the value of k.

My question is, how would I even start this kind of problem? Again, thanks for the help!

Hello and welcome to MHB! :D

I would use the remainder theorem to state:

$$p\left(\frac{4-i}{2} \right)=5+i$$

Next, use the definition of $p$:

$$\left(\frac{4-i}{2} \right)^3+k\left(\frac{4-i}{2} \right)+3-2i=5+i$$

Now expand and solve for $k$. You should get the solution given by mente oscura.

I have moved this thread to our Pre-Calculus subforum as it really is not a question typical to a linear algebra course, except perhaps as a preliminary review question, and so I felt it would be better suited here.

 

[PurpleDude]

Thank you guys very much for the quick help! :)

Also, if I have more questions of the same type, should I ask it in this thread or make a new one?

 

[PurpleDude]

Hello.

I only see divide.

\dfrac{x^3+kx+(3-2i)}{2x-(4-i)}=

=\dfrac{x^2}{2}+\dfrac{x}{4}(4-i)+\dfrac{1}{2}[\dfrac{(4-i)^2}{4}+k]+Rest

Rest=\dfrac{1}{2}[\dfrac{(4-i)^2}{4}+k](4-i)+(3-2i)

Therefore:

Rest=\dfrac{1}{2}[\dfrac{(4-i)^2}{4}+k](4-i)+(3-2i)=5+i

\dfrac{(4-i)^3}{8}+\dfrac{k(4-i)}{2}+3-2i=5+i

(4-i)^3+4k(4-i)=16+24i

64-48i+12i^2-i^3+4k(4-i)=16+24i

64-48i-12+i+4k(4-i)=16+24i

4k(4-i)=-36+71i

Let \ k=a+bi

4(a+bi)(4-i)=-36+71i

16a-4ai+16bi-4bi^2=-36+71i

16a-4ai+16bi+4b=-36+71i

Therefore:

16a+4b=-36

-4ai+16bi=71i \rightarrow{}-4a+16b=71

a=-\dfrac{215}{68}

b=\dfrac{62}{17}

Solution:

k=- \dfrac{215}{68}+\dfrac{62}{17}i

I hope to not have me wrong calculations.

Regards.

I've been trying to do the calculations myself but how did you get 16 + 24i?
The question I am asking is referring to line 7.

Nevermind, I figured it out :p

 

[Sudharaka]

Thank you guys very much for the quick help! :)

Also, if I have more questions of the same type, should I ask it in this thread or make a new one?

Hi PurpleDude,

Welcome to MHB! :)

Rule No. 8 answers your question,

Do not ask more than two questions in a thread or post. It is better for forum organization, and better for you to get your questions answered in a more timely manner, if you start new threads as necessary for remaining questions. E.g., if you have five questions, post four of them in two threads (two questions each thread) and start a new thread for the remaining one. And if the question has more than two parts to it, it is best to post only that question and its parts in one thread and start a new thread for other questions.

You can find the list of rules http://mathhelpboards.com/rules/. And selecting "Explained Rules" from the drop down menu at the top of that page will give you more information.

 

[mente oscura]

I've been trying to do the calculations myself but how did you get 16 + 24i?
The question I am asking is referring to line 7.

Nevermind, I figured it out :p

Hello.

\dfrac{(4−i)^3}{8}+\dfrac{k(4−i)}{2}+3−2i=5+i

\dfrac{(4−i)^3}{8}+\dfrac{k(4−i)}{2}=2+3i

multiplying everything by 8:

(4−i)^3+4k(4−i)=16+24i

Regards.

 

[PurpleDude]

Okay, so only two questions per thread.

My last question is as follows:

Find all values of k (which may be complex numbers) such that kx - (k + 1) is a factor of the polynomial x² + (2 + 3i)x - (17 + i)

Thanks for the help :)

 

[Prove It]

Okay, so only two questions per thread.

My last question is as follows:

Find all values of k (which may be complex numbers) such that kx - (k + 1) is a factor of the polynomial x² + (2 + 3i)x - (17 + i)

Thanks for the help :)

Your polynomial is \displaystyle \begin{align*} P(x) = x^2 + \left( 2 + 3i \right) x - \left( 17 + i \right) \end{align*}.

If \displaystyle \begin{align*} k\,x - \left( k + 1 \right) \end{align*} is be a factor, then \displaystyle \begin{align*} P \left( \frac{k + 1}{k} \right) = 0 \end{align*}. Can you go from here?

 

[MarkFL]

Okay, so only two questions per thread...

This rule refers to two question maximum in the initial post. We prefer that additional questions, even if only one question was asked up front, not be tagged on to the end of an existing thread. Otherwise the thread can become confusing and difficult to follow (follow-up questions are OK).

 

[PurpleDude]

Your polynomial is \displaystyle \begin{align*} P(x) = x^2 + \left( 2 + 3i \right) x - \left( 17 + i \right) \end{align*}.

If \displaystyle \begin{align*} k\,x - \left( k + 1 \right) \end{align*} is be a factor, then \displaystyle \begin{align*} P \left( \frac{k + 1}{k} \right) = 0 \end{align*}. Can you go from here?

I expanded everything and what I get is:

-13k² + 8k + 12ik² + 24ik + 4 + 12i + 9i²k² +18ik + 9i²

and this whole thing is over k²
:(

 

[MarkFL]

You need only set the numerator to zero, and I would group on the real and imaginary terms such that you have the form:

$$a+bi=0$$

What restrictions then do we have on $a$ and $b$?

 

[PurpleDude]

You need only set the numerator to zero, and I would group on the real and imaginary terms such that you have the form:

$$a+bi=0$$

What restrictions then do we have on $a$ and $b$?

I'm not too sure I understand what you mean MarkFL.

I have substituted (k + 1)/k into each "x" in the polynomial. The only path forward I see is to expand and simplify.

 

[MarkFL]

I'm not too sure I understand what you mean MarkFL.

I have substituted (k + 1)/k into each "x" in the polynomial. The only path forward I see is to expand and simplify.

The first thing I would do in your expression is rewrite $i^2$ as $-1$. Now you will have some terms that have $i$ as a factor and some that do not. Group all the real terms together and all the imaginary terms together, so that you then have an expression of the form:

$$a+bi=0$$

where $a$ and $b$ are both functions of $k$. Now, what must be true of $a$ and $b$?

 

[PurpleDude]

The first thing I would do in your expression is rewrite $i^2$ as $-1$. Now you will have some terms that have $i$ as a factor and some that do not. Group all the real terms together and all the imaginary terms together, so that you then have an expression of the form:

$$a+bi=0$$

where $a$ and $b$ are both functions of $k$. Now, what must be true of $a$ and $b$?

Oh, okay I think I'm on the right path. I have grouped the terms and I have something that looks like this:

-14k² + 4k + 1 + 2ik² + 3ik = 0

This is the closest I can get to that experession. Also, what do you mean by what must be true of $a$ and $b$?

 

[MarkFL]

I have not checked your work, but my next step would be to write:

$$\left(-14k^2+4k+1 \right)+\left(2k^2+3k \right)i=0$$

Now if a complex number is equal to zero, that is if:

$$a+bi=0$$

Then it follows that we must have:

$$a=b=0$$

And this implies:

$$a\pm b=0$$

 

[PurpleDude]

I have not checked your work, but my next step would be to write:

$$\left(-14k^2+4k+1 \right)+\left(2k^2+3k \right)i=0$$

Now if a complex number is equal to zero, that is if:

$$a+bi=0$$

Then it follows that we must have:

$$a=b=0$$

And this implies:

$$a\pm b=0$$

Okay! I think I got it, thanks for taking the time to help me out! :)