[calvinuk]

Just been doing some C1 revision, and i've come across something I just cannot do for some reason.

The Curve C with equation passes through the point and

a.) Use integration to find .

b.) Show that .

So I can integrate it, or at least I think I can, but I cannot fathom how to do part B.

Hopefully it's just tiredness, anyone care to put me out of my misery?

[GHOSH-5]

a) well you have f'(x) = 3(x-1)(x+1)

try expanding it and integrating it to find f(x), remembering how integration works:



b) once you've done the integration and found the constant, try factorising it.

[calvinuk]

Yeah, for that I got:



I then found C to be 2.

[GHOSH-5]

Originally Posted by calvinuk

Yeah, for that I got:



I then found C to be 2.

yep that's right, so

Now we need to factorise this. try finding out values of f(x) for some small integers 'x', like x = 2, x = -2, x = 1 etc, and see if you can make it equal zero. Then use factor theorem:



And in this way, try to factorise f(x).

Spoiler:

of course looking at the question, x=1,-2 automatically gives us a root, and then we can factorise the cubic as:

for some integers a, b and c

[calvinuk]

Hmm, I think I understand now.

So when

[GHOSH-5]

Originally Posted by calvinuk

Hmm, I think I understand now.

So when

well not quite. you've got the factorisation, but the idea of computing f(1), f(2) etc was just to actually find for which values of x does f(x) = 0, and then by the factor theorem, this provides us with a way to factorise the expression.

is true for all x, not just when x = 1, or x = -2.

Spoiler:

perhaps a further example will make it clear:
we want to factorise

so we can try some small integer values of x, and see if we find a point when g(x) = 0


So we have already found a value of x when g(x) = 0.

Now factor theorem states that: g(a) = 0 if and only if (x-a) is a factor of g(x)
Here a = 1

Therefore we know that (x-1) is a factor of g(x)

We could factorise g(x) on inspection now, or use long division or we could try some more values of x:

so (x-2) is not a factor.



This implies that (x+1) is a factor of g(x)



This implies that (x+2) is a factor of g(x)

Being a cubic, there must be at most three factors, and as we have found three we can say that:



to check, expanding the factors on the right yields the cubic equation, so it has been correctly factorise.

You might like to try and factorise:

hopefully that will help somewhat.