Recent content by Celestion

I'll practice it next week and see how it goes.

Yep, thanks very much, I reckon that's how they did it in the worked solution. One of the examples in the PDF (number 5) is almost the same question as my original post.

Yep, that's to find the real integer factors which in this case is x-1, i.e. that p(1)=0. They are saying that assuming you already know your divisor is a factor, you can use this "method of inspection" to find the quotient. The rest of the explanation is in the PDF I linked to a couple of posts...

I'm quite happy doing the long division, though for tutoring other students I want to know all the "tricks" that may apply

I'll look at it properly later but now I highly suspect they're referring to the "Method of inspection" which was in a pdf that I just discovered, from a third party "tips and tricks" course for this subject. It begins like this "The method of inspection: If you have a known factor, say from...

Check out the "method of inspection" on page 6. I just found this but I haven't got time to look at it properly until Monday. I'm not sure how much quicker it would be but it's the first I've seen of a section called "Avoiding polynomial long division" that begins on p5. This isn't an official...

There is a part in the question above that where some things are done with α, β, and γ (using the results like α+β+γ=-b/a) but none of that resembles anything that I can see how to pick the quadratic out of, there's nothing that divides the polynomial and none of the numbers for any of the...

Yes it's in English. The problem was in the Australian NSW higher school certificate extension 2 maths exam for 2001. The worked solution is in a book by Coroneos publications. What it says literally is "Thus P(x) = (x-1)Q(x) where Q(x) = x2 -2x +2 at sight, or see note." Then it goes on to...

I looked up Horner's rule and it seems to mean rewriting the polynomial x3 - 3x2 + 4x - 2 = 0 as ((x-3)x + 4)x - 2 but I don't see how that helps to divide it by x-1?

In the worked solution, it doesn't mention factoring the quadratic to give 1-i and 1+i in your head, they use completing the square.

The first term of x2 in x2 - 2x + 2 can be seen easily in your head. The last term of 2 also looks kind of obvious, is it generally true that the constant terms have to evenly divide? (i.e. that the -2 in the original cubic divided by the -1 in x-1 gives 2). This seems to make sense assuming...

Thanks everyone. I'm not familiar with Horner's rule so I'll check that out. I can do basic factorisations like x2 +5x +6 in my head but that's about it.

Homework Statement The question was to find the roots of x3 - 3x2 + 4x - 2 = 0 Homework Equations The first root is found by the factor theorem, substituting x=1 into the polynomial gives 0 therefore x=1 is one root and (x-1) is a factor. The Attempt at a Solution In the worked solution...

Thanks Andrew, that's great, it makes perfect sense. I think their/your solution is more elegant than mine :) now that I understand it

Homework Statement I can do the question, but in a different way to the worked solution which I don't understand. So my question is can anyone explain the worked solution which is in point 3 below. The question was to show there is exactly one zero to the function f(x) = Ax^3 - Ax + 1, with...