# Problem with simplifying fractions

1. Sep 17, 2006

### jcheema

Problem with factorising

Hi, the question is to differentiate the following equation with respect to x.

x^4(3x-1)^3

Using the product rule i think i'v partially completed this to

x^4(3(3x-1)^2) + (3x-1)^3(4x^3)

I'm now required to simplify this - which leaves me completely stumped. There is an answer in the back of the book which i simply cannot get to, could i please hear some of your solutions to this!

James

Edit:Wrong title

Last edited: Sep 17, 2006
2. Sep 17, 2006

### danago

There is a slight error in that derivative you have come to.

$$\frac{d}{{dx}}((3x - 1)^3 ) = 9(3x - 1)^2$$

You seem to have put 3 instead of 9

Other than that, it looks fine to me. After changing it, try taking out a factor of $$x^3 (3x - 1)^2$$

3. Sep 17, 2006

### jcheema

So it would now become

x^4(9(3x-1)^2) + (3x-1)^3(4x^3) - i see my mistake there, my only problem now is how to simplify this further. How would i go about taking the factor out?

4. Sep 17, 2006

### danago

I replied to your private message explaining everything. If the equations i wrote in that dont show up, ill post it here :)

5. Sep 17, 2006

### danago

Well it seems that the LaTeX wouldnt work in private messages. Here is exacly what i messages you:

6. Sep 17, 2006

### jcheema

Excellent description there :) Beats my textbook! However i'm now just struggling to see how

$$9x^4 a^2 + 4x^3 a^3 = x^3 a^2 (9x + 4a)$$

As i'd work out x^3*9x, which would give you the 9x^4. But i'm not really sure how it'd then multiply out to give you the other terms, apologies as my maths skills are not the best and i ask you to please have patience with me!

7. Sep 17, 2006

### danago

ok. When you are factorizing things, essentially, you are dividing through by something, and then multiplying by that same thing again.

Look at:
$$9x^4 a^2 + 4x^3 a^3$$

You can divide the whole expression by x, to get:

$$9x^3 a^2 + 4x^2 a^3$$

But you cant just divide like that, because you are then changing the value of the expression. So to "cancel out" this division process, you then do the opposite of that division by x, which would be to multiply by x.

$$x(9x^3 a^2 + 4x^2 a^3 )$$

What about now if i divided by 'xa'.

$$9x^3 a + 4x^2 a^2$$

To make that division process valid, i must now multiply again by 'xa', and i get:

$$xa(9x^3 a + 4x^2 a^2)$$

What about if we do the same now, but instead divide by $$x^3 a^2$$.

$$\frac{{9x^4 a^2 + 4x^3 a^3 }}{{x^3 a^2 }} = 9x + 4a$$

But dividing by that completely changed the value of the expression. So to make that division valid, we now have to multiply the whole thing by what we divided it by:

$$x^3 a^2 (9x + 4a)$$

And thats where the $$9x^4 a^2 + 4x^3 a^3 = x^3 a^2 (9x + 4a)$$ came from.

Did that make any sense to you?

8. Sep 17, 2006

### jcheema

That makes perfect sense yes, thanks very much for your time - you're a credit to the forums

9. Sep 17, 2006

### danago

No problems :) Glad to be of assistance :)

Im glad you understand it now :)