# Thread: finding values of variables in a equation

1. ## finding values of variables in a equation, polynomial long division

I found this question on the internet while testing my knowledge and skills of math. This question has been use as part of the Flemish entrance exam for the study medicine (along with physics, biology, chemistry and psychological tests).
When $\displaystyle x^4 + 4x^3 + 6px^2 + 4qx + r$ is divisable by $\displaystyle x^3 + 3x^2 + 9x + 3$ then what equals $\displaystyle p(q+r)$?
A. 12
B. 15
C. 18
D. 21
The correct answer is A. Can anyone please explain to me what approach you use to find this answer? I have absolutely no clue how to solve this. Source of the question: http://www.ond.vlaanderen.be/toelati...n_wiskunde.pdf
It's the fourth question.

2. Originally Posted by Sebastian de Vries
I found this question on the internet while testing my knowledge and skills of math. This question has been use as part of the Flemish entrance exam for the study medicine (along with physics, biology, chemistry and psychological tests).
When $\displaystyle x^4 + 4x^3 + 6px^2 + 4qx + r$ is divisable by $\displaystyle x^3 + 3x^2 + 9x + 3$ then what equals $\displaystyle p(q+r)$?
A. 12
B. 15
C. 18
D. 21
The correct answer is A. Can anyone please explain to me what approach you use to find this answer? I have absolutely no clue how to solve this. Source of the question: http://www.ond.vlaanderen.be/toelati...n_wiskunde.pdf
It's the fourth question.
do you know how to do long division (or synthetic division) of polynomials?

if you do the division, you will end up with $\displaystyle (6p - 12)x^2 + (4q - 12)x + (r - 3)$ as the remainder.

But since we are told the first polynomial is divisible by the second, it means the remainder has to be zero (why?)

thus we need: $\displaystyle 6p - 12 = 0$, $\displaystyle 4q - 12 = 0$ and $\displaystyle r - 3 = 0$

this means, $\displaystyle \boxed{p = 2}, ~ \boxed{q = 3}, \text{ and } \boxed{r = 3}$

plugging these values into $\displaystyle p(q + r)$ we find that the answer is 12, which is choice A

Originally Posted by Jhevon
do you know how to do long division (or synthetic division) of polynomials?
Oddly enough not.
I am aquinatanced with the ordinary long division* but I've never learned how to apply this on polynomials. I followed my high school in the Netherlands where I studied math but since this math was of a lower level than the math of the Flemish (Belgium) high schools with sometimes a different system of notation I have some difficulty with it (at the university they assume that you have a solid basis of algebra, calculus and even some linear algebra before you enter).
I found a link on wikipedia (Polynomial long division - Wikipedia, the free encyclopedia ), I have only had a superficial look at it but I think I won't have too much difficulty to follow it. I'll let it know tomorrow after having read it completely.

*At the Dutch grammar school they quit with teaching the long division (the ordinary long division with numbers) in the eighties. The reason for it? The weaker pupils had difficulty with this correcter system so the teachters collectively replaced the long division by a different system wich was easier to use but gave less insight.
By the way, I hope that it's not too obvious that English isn't a native language for me.

4. Originally Posted by Sebastian de Vries
I am aquinatanced with the ordinary long division* but I've never learned how to apply this on polynomials. I followed my high school in the Netherlands where I studied math but since this math was of a lower level than the math of the Flemish (Belgium) high schools with sometimes a different system of notation I have some difficulty with it (at the university they assume that you have a solid basis of algebra, calculus and even some linear algebra before you enter).
I found a link on wikipedia (Polynomial long division - Wikipedia, the free encyclopedia ), I have only had a superficial look at it but I think I won't have too much difficulty to follow it. I'll let it know tomorrow after having read it completely.
also, have a look at:

Polynomial Long Division

Polynomial Long Division

Synthetic Division

Synthetic Division -- from Wolfram MathWorld

*At the Dutch grammar school they quit with teaching the long division (the ordinary long division with numbers) in the eighties. The reason for it? The weaker pupils had difficulty with this correcter system so the teachters collectively replaced the long division by a different system wich was easier to use but gave less insight.
By the way, I hope that it's not too obvious that English isn't a native language for me.
your english is fine. better than many native english speakers even

5. All right, I get it. It explains why I couldn't follow some steps with physics when they rewrote functions. It seemed complicated but it is quite simple.

$\displaystyle \frac{3x^4+4x^3+6px^2+4qx+r}{x^3+3x^2+9x+3}=x+1+\f rac{(6p-12)x^2+(4q-12)x+r-3}{x^3+3x^2+9x+3}$

The remainder has to be zero because the definition of divisionable is that there has to be no reamainder.
So it turns out that I had two problems with this question:
I hadn't learned how to execute a polynomial division and I didn't know the definition of divisionable. This definition is, for me, counterintuitive. I assumed that every fraction wich is a rational number (domain of $\displaystyle \mathbb{Q}$) would be defined as divisionable.
So it shows that assumptions are the mothers of many srew ups.

6. Originally Posted by Sebastian de Vries
All right, I get it. It explains why I couldn't follow some steps with physics when they rewrote functions. It seemed complicated but it is quite simple.

$\displaystyle \frac{3x^4+4x^3+6px^2+4qx+r}{x^3+3x^2+9x+3}=x+1+\f rac{(6p-12)x^2+(4q-12)x+r-3}{x^3+3x^2+9x+3}$

The remainder has to be zero because the definition of divisionable is that there has to be no reamainder.
So it turns out that I had two problems with this question:
I hadn't learned how to execute a polynomial division and I didn't know the definition of divisionable. This definition is, for me, counterintuitive. I assumed that every fraction wich is a rational number (domain of $\displaystyle \mathbb{Q}$) would be defined as divisionable.
yup, that's it.

our answer came by setting the numerator of the remainder term to zero. since it had to be zero regardless of the value of x, we needed the coefficients to be zero

So it shows that assumptions are the mothers of many srew ups.
well, we all know what ass-u-me stands for...