MHB 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:

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

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

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

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

Now, adding the two equation, we find:

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

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

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

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:

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

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

We are then asked to find the expression for $$ax^2+bx+c$$ in

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

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

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

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

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

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

It is obvious that we now have $$b-a=1$$ and $$c-a=5$$.

Solving $$a-b+c=-2$$, $$b-a=1$$ and $$c-a=5$$ for the values of a, b and c and we get

$$a=-6$$, $$b=-5$$ and $$c=-1$$.

 

[MarkFL]

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

 

[gabriel1]

...

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

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

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,
$$f(x)=(Q_3(x)(x+1)a)(x^2+x+1)+(b-a)x+c-a$$ should be read as
$$f(x)=[(Q_3(x)(x+1))+a](x^2+x+1)+(b-a)x+c-a$$ instead. Thanks for fixing it for me and glad to help out!