Dividable polynomes

[Function]

Hello

 

In Belgium, you have to pass an exam in order to commence Medicine at university. Here's an example question:

 

Be [math]8x^4+10x^3-7px^2-5qx+9r[/math] dividable by [math]4x^3+7x^2-21x-18[/math], then [math]p+q+r=?[/math]

• 12
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• 15

In order to do this, I divided the first polynome by the second, resulting in quotient [math]2x-1[/math] and rest [math](49-7p)x^2+(15-5q)x-(9r-18)[/math] and as the first polynome is dividable by the second one, [math](49-7p)x^2+(15-5q)x-(9r-18)=0[/math]

 

Can someone help me on solving this problem?

 

Thanks.

 

Function

Edited January 22, 2014 by Function

[studiot]

I make the constant term of your quadratic to be (9r-18).

[Function]

Oversaw that one. My bad. Any idea on how to solve this problem?

[studiot]

Are you familiar with the remainder theorem?

[Function]

Yes. I've seen it well.. a long time ago at school. I'm afraid, though, I can't remember it. I'll Google it!

 

Is it this: the rest (or remainder, whatever it is) of a division of [math]f(x)[/math] by [math](x-a)[/math] is [math]f(a)[/math]?

Edited January 22, 2014 by Function

[studiot]

Yes that's it.

 

But this theorem leads to a number of results.

 

Since your quadratic is one of the factors of the original quartic,(x=k, x=l, x = m, x=n ), any solution of the quadratic is also a solution of the quartic.

So

 

this gives you additional equations to connect p, q and r.

 

Further the two solutions (x=k and x=l) of the quadratic ax² + bx +c have the folowing properties

 

(k+l) = -b/a (note the negative)

 

kl = c/a

 

Which also supply equations connecting p, q and r

Edited January 22, 2014 by studiot

[studiot]

My solution makes p+q+r = 12.

 

I think the quadratic is a red herring that you do not need.

 

Although the answer comes out easily in a few lines, I'm amazed they consider it useful for medical studies.

Edited January 23, 2014 by studiot

[studiot]

Here is a worked solution

 

 

 

Consider

8x⁴+10x³-7px²-5qx+9r

 

This may be rewritten

 

(4x³+7x²-21x-18)(2x-1) +R

 

Where R is a polynomial in x.

 

We are told that the cubic divides the quartic exactly so R = 0

 

thus multiplying out the terms in brackets we get

 

8x⁴+10x³-49x²-15x+18

 

Comparing this with the original quartic and equating coefficients we find

 

9r = 18 : r = 2

 

-5q = -15 : q = 3

 

-7p = -49 : p = 7

 

Thus (p + q + r) = 12

 

 

 

Edited January 23, 2014 by studiot

[Function]

Thank you very much!