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# Remainder Theorem watch

1. Can someone help me as to where to start? I think it may be simultaneous☹️
2. f(1)=f(-1/2) i think

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3. (Original post by JaisulNaikJN)
f(1)=f(-1/2) i think

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I'm not sure if that's necessarily true.

(Original post by ozmo19)

Can someone help me as to where to start? I think it may be simultaneous☹️
I would suggest long division but there may be a quicker method.
4. Would yuo not do polynomial division...? Thats what I would do
5. (Original post by ozmo19)

Can someone help me as to where to start? I think it may be simultaneous☹️
As JaisulNaikJN said, "f(1)=f(-1/2)"

q will cancel and you're left with an equation solely in p.
6. I think p=18?
7. Okay the steps to do this:

1) Do a long polynomial division #1:
f(x) / (x-1) --> You'll get a remainder, we shall call #1

2) Do a second polynomial division #1
f(x) / (2x +1) --> You'll get a remainder, we shall call #2

3) Set remainder #1 = remainder #2. From this, you should be able to work out what p is. You might need to re-arrange the remainders in the form q= p(x).

Video help for steps 1 and 2:

Website help for steps 1 and 2:
http://www.purplemath.com/modules/polydiv2.htm

As to why polynomial division is important to real world applications, well it may help us simplify equations down, potentially so that we can find solutions more easily. Maybe a mathematician can expand more on some reasons!
8. (Original post by SeanFM)
I'm not sure if that's necessarily true.

I would suggest long division but there may be a quicker method.
That is true because the Remainder Theorem states that f(x) divided by (x-p) gives the remainder f(p). You can do the same with both of those and you get the same remainder, therefore both are equal. you can then use simultaneous equations to work out your answer

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