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# Polynomial C2 Proof watch

1. Show that if (x -p) is a factor of f(x) then f(p) = 0.

So common sense told me that it would look something like this: f(x) = (x - p)

Then, f(p) = (p -p) so f(p) = 0

But having looked at the answer it states that f(p) = 0 x g(p)

I have missed some kind of logical step, but i cant figure out what g(p) actually is?

Thanks!!
2. (Original post by JamesNeedHelp2)
Show that if (x -p) is a factor of f(x) then f(p) = 0.

So common sense told me that it would look something like this: f(x) = (x - p)

Then, f(p) = (p -p) so f(p) = 0

But having looked at the answer it states that f(p) = 0 x g(p)

I have missed some kind of logical step, but i cant figure out what g(p) actually is?

Thanks!!
(x-p) is a factor, but not necessarily the whole of the polynomial.

g(x) is what's left when you take out (x-p)

E.g.

Let f(x) = x^4-px^3

We can factorise this to (x-p)x^3

Our g(x) in this case is x^3
3. (Original post by ghostwalker)
(x-p) is a factor, but not necessarily the whole of the polynomial.

g(x) is what's left when you take out (x-p)

E.g.

Let f(x) = x^4-px^3

We can factorise this to (x-p)x^3

Our g(x) in this case is x^3
Isnt g(x) in simple terms then just the polynomial equation?
4. (Original post by JamesNeedHelp2)
Isnt g(x) in simple terms then just the polynomial equation?
No

If you have a polynomial f(x) and it can be factorised ... This means that you can split it into two or more things that can be multiplied together

So f(x) = (x-p)g(x)
5. (Original post by JamesNeedHelp2)
Isnt g(x) in simple terms then just the polynomial equation?
No.

The statement "f(x) is divisible by x-p" means the same as "we can write f(x) = (x - p)g(x) where g(x) is another polynomial with a lower degree than f(x)".

x^2 - 1 is divisible by x - 1 because x^2 - 1 = (x - 1)(x + 1) but x - 1 isn't "the whole of the original polynomial".
6. (Original post by TenOfThem)
No

If you have a polynomial f(x) and it can be factorised ... This means that you can split it into two or more things that can be multiplied together

So f(x) = (x-p)g(x)
f(x) = (x-p) x g(x)

f(p) = (p-p) x g(p)

f(p) = (0) x g(p)

f(p) = 0

so what you guys are saying is that g(x) is one of the factors that when multiplied by another factor in this case (x-p) equates to f(x)?
7. (Original post by JamesNeedHelp2)
f(x) = (x-p) x g(x)

f(p) = (p-p) x g(x)

f(p) = (0) x g(x)

f(p) = 0

so what you guys are saying is that g(x) is one of the factors that when multiplied by another factor in this case (x-p) equates to f(x)?
Yes ... Apart from ... In your working when you sub p in you should have g(p) not g(x)

Using x for multiply is unnecessary and confuses the issue
8. (Original post by TenOfThem)
Yes ... Apart from ... In your working when you sub p in you should have g(p) not g(x)

Using x for multiply is unnecessary and confuses the issue
Yeah, my bad. Cant really rep you, as it wont allow me. Would be awesome if there was some kind of loop hole in the rep system, so that i could rep you to infinity hah
9. (Original post by davros)
No.

The statement "f(x) is divisible by x-p" means the same as "we can write f(x) = (x - p)g(x) where g(x) is another polynomial with a lower degree than f(x)".

x^2 - 1 is divisible by x - 1 because x^2 - 1 = (x - 1)(x + 1) but x - 1 isn't "the whole of the original polynomial".
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