-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
prelim question (1 Viewer)
- Thread starter ta1g
- Start date
P
pLuvia
Guest
lol.. prob solve it.. haha or differentiate??
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
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)
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)
Trev
stix
Forgot to cancel there...Loz_metalhead said:
x^2(x+3)-9(x+3)/(x+3)(x+3) = (x^2-9)/(x+3) = x-3.
haack_m
Member
its to early to be thinking maths,
