1. ## Confused by Factorising, perculiar answers

I have an exam tomorrow and have just spotted a flaw in my brain.

I am using a revision book with all the answers.

3 questions:
This is what I'm confused about.

$\displaystyle x/p = p+c$
so
$\displaystyle x=p^2+pc$

But WHERE did the extra p come from. Surely thers only 1 p why is there now two?
$\displaystyle x = p^2+c$

Next I've got:
$\displaystyle 5-fx=3x+p$
How does this =
$\displaystyle x=(5-p)/(3+f)$

I simply don't understand.
I got stuck at:
$\displaystyle x=(5-fx-p)/3$

Finally I have a mega confusion of another letter just dissapearing.

$\displaystyle 2x=px+q$

so

$\displaystyle 2x-px=q$
it then says "factorise to"
$\displaystyle (2-p)x=q$
Where's the $\displaystyle x$ gone!? I see what they've done, but WHY!? again.

Will appreciate any help or nudges in the right direction. I also dont know which side of an equation something should go, if i -5 from both sides, do I put the -5 in front or behind the rest of the equation?

2. $\displaystyle \displaystyle \frac{x}{p}= p+c$

To isolate $\displaystyle x$ you mulitply both sides by $\displaystyle p$ giving,

$\displaystyle \displaystyle x=p( p+c)$

Then this p distributes through the brackets giving

$\displaystyle \displaystyle x=p\times p+p\times c$

$\displaystyle \displaystyle x=p^2 +pc$

I think there was a second question in your original post, the latex did not output clearly.

3. Originally Posted by sxsniper

Finally I have a mega confusion of another letter just dissapearing.

$\displaystyle 2x=px+q$

so

$\displaystyle 2x-px=q$
it then says "factorise to"
$\displaystyle (2-p)\mathbf{x}=q$
Where's the $\displaystyle x$ gone!? I see what they've done, but WHY!? again.

Will appreciate any help or nudges in the right direction. I also dont know which side of an equation something should go, if i -5 from both sides, do I put the -5 in front or behind the rest of the equation?
It's right there, where, with any luck, it's bold

4. Originally Posted by sxsniper
Next I've got:
$\displaystyle 5-fx=3x+p$
How does this =
$\displaystyle x=(5-p)/(3+f)$
$\displaystyle \displaystyle 5-fx=3x+p$

$\displaystyle \displaystyle 5=3x+fx+p$

$\displaystyle \displaystyle 5-p=3x+fx$

Factor out $\displaystyle x$

$\displaystyle \displaystyle 5-p=x(3+f)$

$\displaystyle \displaystyle x= \frac{5-p}{3+f}$

5. Although I know what Factorise means.

Everything always just says "factorise" and then gives the factorised answer, this is my main problem. It's how you go about it step by step it. The sentance "Factorise out x" is understandable, but the process isn't. All the books im reading just keep saying "factorise" and bang there's the answer. Which is really not that helpful for me.

I can factorise most stuff, but the stuff I've highlighted here I simply don't know how to go about it.

*EDIT* Ah nevermind it's 2am in the morning, I think you did explain it. Hmm I'm going to have to read it in the morning. I'm passing out here.

6. Originally Posted by pickslides
$\displaystyle \displaystyle 5-p=3x+fx$

factor out $\displaystyle x$

$\displaystyle \displaystyle 5-p=x(3+f)$
that bit!

7. Ah I see.

$\displaystyle x*3 = 3x$ and $\displaystyle x*f = fx$
Thanks.

8. this post was made redundant - see above

9. Originally Posted by e^(i*pi)
It's right there, where, with any luck, it's bold
That was ridiculously unhelpful. However, due my recent breakthrough, it was somewhat helpful .

I'm still a little confused about which side of an expression to put the subject when you subtract it.

10. I almost feel immortal when the penny drops.