Gabriel's question at Yahoo Answers regarding polynomial division and remainders

[MarkFL]

Here is the question:

The remainder of f(x)/(x^2+x+1) and f(x)/[(x+1)^2] are x+5 and x-1 respectively.?

What is the remainder of f(x)/[(x^2+x+1)(x+1)]?

Here is a link to the question:

The remainder of f(x)/(x^2+x+1) and f(x)/[(x+1)^2] are x+5 and x-1 respectively.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Gabriel,

We are told:

\(\displaystyle \frac{f(x)}{x^2+x+1}=Q_1(x)+(x+5)\)

\(\displaystyle \frac{f(x)}{(x+1)^2}=Q_2(x)+(x-1)\)

If we multiply the second equation by $x+1$, we have:

\(\displaystyle \frac{f(x)}{x+1}=Q_2(x)(x+1)+(x^2-1)\)

Now, adding the two equation, we find:

\(\displaystyle f(x)\left(\frac{x^2+2x+2}{(x+1)(x^2+x+1)} \right)=Q_1(x)+Q_2(x)(x+1)+(x^2+x+4)\)

Dividing through by $x^2+2x+2$ and then defining \(\displaystyle Q_3(x)\equiv\frac{Q_1(x)+Q_2(x)(x+1)}{x^2+2x+2}\), we obtain:

\(\displaystyle \frac{f(x)}{(x+1)(x^2+x+1)}=Q_3(x)+\frac{x^2+x+4}{x^2+2x+2}\)

To Gabriel and any other visitors viewing this topic, I invite and encourage you to register and post any other algebra questions in our http://www.mathhelpboards.com/f2/ forum.

Best Regards,

Mark.

 

[gabriel1]

Thank you for the help~

However, the first two lines should be...
\[
\frac{f(x)}{x^2+x+1}=Q_1(x)+\frac{x+5}{x^2+x+1}\\
\frac{f(x)}{(x+1)^2}=Q_2(x)+\frac{x-1}{(x+1)^2}
\]

and the given answer is \[-6x^2-5x-1\]

 

[anemone]

We are told:

\(\displaystyle f(x)=Q_1(x)(x^2+x+1)+(x+5)\) (*)

\(\displaystyle f(x)=Q_2(x)(x+1)^2+(x-1)\) (**)

We are then asked to find the expression for \(\displaystyle ax^2+bx+c\) in

\(\displaystyle f(x)=Q_3(x)(x^2+x+1)(x+1)+(ax^2+bx+c)\) (***)

First, we can use the fact that \(\displaystyle f(-1)=-1-1=-2\) to the equation (***) and obtain:

\(\displaystyle f(-1)=-2=a-b+c\) or simply \(\displaystyle a-b+c=-2\)

But we know that we can rewrite the equation (***) so that it takes the form in (**) in the following manner:

\(\displaystyle f(x)=Q_3(x)(x^2+x+1)(x+1)+a(x^2+x+1)-ax-a+bx+c\)

\(\displaystyle f(x)=(Q_3(x)(x+1)a)(x^2+x+1)+(b-a)x+c-a\)

It is obvious that we now have \(\displaystyle b-a=1\) and \(\displaystyle c-a=5\).

Solving \(\displaystyle a-b+c=-2\), \(\displaystyle b-a=1\) and \(\displaystyle c-a=5\) for the values of a, b and c and we get

\(\displaystyle a=-6\), \(\displaystyle b=-5\) and \(\displaystyle c=-1\).

 

[MarkFL]

Yeah, I got in a hurry and botched that one big time. (Giggle) Thank you anemone! (Cool)

 

[gabriel1]

...

\(\displaystyle f(x)=Q_3(x)(x^2+x+1)(x+1)+a(x^2+x+1)-ax-a+bx+c\)

\(\displaystyle f(x)=(Q_3(x)(x+1)a)(x^2+x+1)+(b-a)x+c-a\)

It is obvious that

...

I think the highlighted line should be
\[ f(x)=(x^2+x+1)((Q_3(x))(x+1)+a)+(b-a)x+c-a \]Thank you anemone for solving the problem!(Handshake) and thank you Mark for the help as well (Smile)

 

[anemone]

I think the highlighted line should be
\[ f(x)=(x^2+x+1)((Q_3(x))(x+1)+a)+(b-a)x+c-a \]

Ops... I was too in a hurry and yes,
\(\displaystyle f(x)=(Q_3(x)(x+1)a)(x^2+x+1)+(b-a)x+c-a\) should be read as
\(\displaystyle f(x)=[(Q_3(x)(x+1))+a](x^2+x+1)+(b-a)x+c-a\) instead. Thanks for fixing it for me and glad to help out!