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prelim question (1 Viewer)

ta1g

Member
Joined
Aug 11, 2005
Messages
30
Can someone show me how to do this question?

x^3+3x^2-9x-27/x^2+6x+9

thanks
 

joey_prince42

New Member
Joined
Apr 4, 2005
Messages
6
Gender
Male
HSC
2005
Hey...try doin this

Hello, I think this may help.

Firstly, I think this question is a bit unfair considering polynomials, e.g. p(x) = ax^n+a2x^n-1+...c , are in the 3/4 U maths courses. Nevertheless I will offer a variety of solutions that may assist you and that you can understand.

Now with x^3+3x^2-9x-27/x^2+6x+9

x^3 + 3^x2-9x-27 = (x-3)(x+3)^2

x^2 + 6x + 9 = (x+3)^2

Therefore the answer is x-3

Of course to come up with such an answer is problematic as you have to be able to find the correct factors. I'll offer some tips that may help without confusing you.

i) In the 2U course it is very unlikely that for these type of questions that they will give you polynomials with irrational or non-integral roots.

ii) As well, as I did for the cubic numerator, you may want to use the factors of the constant to find the other factor/root - i.e in this case -27 : which as factors 3,-3,9,-27,1 and so on. To test if the factor is the root i.e. if x=a and a is the factor, then sub in it in like this P(-a) and if this equals zero, then it is a root. And furthermore if you differentiate the polynomial and find that the same root satifies the differianted function then it is a double root, i.e it is counted "twice" like (x-a)^2 - By the way this is 4U maths work in polynomials called multiple roots.

iii) Another way in the 3/4U course is using polynomial division, so it may be worthwhile to look over that if you cant solve it.

Regards,

Joey ;)
 

Loz_metalhead

Member
Joined
Feb 5, 2005
Messages
800
Gender
Female
HSC
2006
I got (x+3)(x-3)

x^3+3x^2-9x-27/(x+3)(x+3)

x^2(x+3)-9(x+3)/(x+3)(x+3)

=x^2-9

=(x+3)(x-3)
 

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