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

Staff member

#### MarkFL

Staff member
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 Pre-Algebra and Algebra forum.

Best Regards,

Mark.

#### gabriel

##### New member
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

##### MHB POTW Director
Staff member
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

Staff member
Yeah, I got in a hurry and botched that one big time. Thank you anemone!

#### gabriel

##### New member
.....

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

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

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!!! and thank you Mark for the help as well

#### anemone

##### MHB POTW Director
Staff member
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!