# Thread: Can you simplify this further?

1. ## Can you simplify this further?

Hi all,

Had an exam recently and I've got a nagging feeling about a particular question.

$\displaystyle f(x) = \frac{1-x^2}{1-x}$ and $\displaystyle g(t) = t^2$

Find $\displaystyle f(g(t))$ and simplify your answer.

So that results in

$\displaystyle \frac{1-(t^2)^2}{1-t^2}$

Which simplifies to

$\displaystyle \frac{1-t^4}{1-t^2}$

Is there any further simplification we can do? I can't see anything but I have this nagging feeling I'm missing something. Possibly further factorisation taking place considering 1 is also $\displaystyle 1^2$ etc leading to some cancelling out.

Did I miss anything?

2. Yes, of course that simplifies. There is a general formula that $\displaystyle a^n- b^n= (a-b)(a^{n-1}+ a^{n-2}b+ \cdot\cdot\cdot+ ab^{n-2}+ b^{n-1}$. In particular
$\displaystyle 1- t^2= (1- t)(1+ t)$
and
$\displaystyle 1- t^4= (1- t)(1+ t+ t^2+ t^3)$

3. Originally Posted by HallsofIvy Yes, of course that simplifies. There is a general formula that $\displaystyle a^n- b^n= (a-b)(a^{n-1}+ a^{n-2}b+ \cdot\cdot\cdot+ ab^{n-2}+ b^{n-1}$. In particular
$\displaystyle 1- t^2= (1- t)(1+ t)$
and
$\displaystyle 1- t^4= (1- t)(1+ t+ t^2+ t^3)$
True, however I personally wouldn't consider

$\displaystyle \frac{1+ t+ t^2+ t^3}{1+ t}$

which that cancels down as a "simpler" answer than the original though, would you?

4. Originally Posted by Peleus Hi all,

Had an exam recently and I've got a nagging feeling about a particular question.

$\displaystyle f(x) = \frac{1-x^2}{1-x}$ and $\displaystyle g(t) = t^2$

Find $\displaystyle f(g(t))$ and simplify your answer.

So that results in

$\displaystyle \frac{1-(t^2)^2}{1-t^2}$

Which simplifies to

$\displaystyle \frac{1-t^4}{1-t^2}$

Is there any further simplification we can do? I can't see anything but I have this nagging feeling I'm missing something. Possibly further factorisation taking place considering 1 is also $\displaystyle 1^2$ etc leading to some cancelling out.

Did I miss anything?
Yes.

$\displaystyle \frac{1-t^4}{1-t^2} = \frac{(1 - t^2)(1 + t^2)}{1 - t^2} = 1 + t^2$, $\displaystyle t \neq \pm 1$.

5. Bother, knew I missed something.

I originally had it as

$\displaystyle = \frac{(1-t^2)^2}{1-t^2}$

$\displaystyle = \frac{(1-t^2)(1-t^2)}{1-t^2}$

$\displaystyle = (1-t^2)$

But realised I wasn't correct because it was never squared, it was $\displaystyle 1-(g(t))^2$ so I scribbled it out and didn't spot the other way.

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