(28 - 7b)/ (b -4) * (1/ ( b + 10)
Are you able to factorise 28-7b [which is the numerator of the first term]?
yes
What did you get when you factorised it?
i don't know how to factorize the problem i kind of tried and got confused what did you get
Come on eb you got this!
You told me that you knew how to factorise: 28-7b. I want to see what you got for just doing that?
\[\frac{ 27-7b}{ b ^{2-}40 }\]
You're getting all the knowledge that you've learnt mixed up. You're confusing yourself even more, when I asked you to factorise 28-7b. Could you do that step for me please?
i forgot how to do that i thought you was talking about subtracting them
If you factorise, you take a common factor for each of the terms. For EXAMPLE: Factorise 4+8x. The factor for the two terms is 4. So we take the 4 out and we get: \[4(1+2x)\] WHen you distibute or expand, you would get the same two terms as before.
One more note- If you take out the common factor, you have would be changing the terms inside the brackets. If you factorise 4+8x and take out 4, you don't go like this: \[4(4+8x)\] That would give you a completely different result. You must account for the factor taken out and you must make it possible to fit everything together. What would be the factor for 28-7b then?
you would be*
7(4-1b)
Well done. you don't need the 1 because 1b=b. It's the same thing. Okay if we also factorise the minus sign out, what would we get then?
So for EXAMPLE, if we take out the minus sign in this example: 4(1-2x)=-4(2x-1), that would be the end result.
okay
Could you show me what you get now when you take out the minus from: 7(4-b)?
7(4b)
Did you read my example properly?
Factorise the minus also means take the minus out.
I don't mean literally take it out. I mean factorise the minus sign out.
Don't you see that if you do that, it wouldn't make sense mathematically if you literally took the minus sign out?
-7(b-4)
@eb3001 YES! Now you have this: \[\frac{-7(b-4)}{b -4} \times \frac{1}{b + 10}\]
Now you can cancel the b-4 with the denominator and numerator.
@eb3001 Come on you're almost done! Just follow along with @Azteck
alright \[-7*\frac{ 1 }{ b+10 }\] so you get this
\[-7* \frac{ -7 }{ 7b+70 }\]
\[\frac{ 7b }{ 7b+70}\]
Oops, you can't multiply the whole thing by -7, just the numerator. If it was -7/-7 then in that cause you could.
\[\frac{ -7b }{ b+10 }\]
Where did that b come from?
Um you were correct when you came to this stage: \[-7\times \frac{1}{b+10}\]
The last step would just be putting the -7 on top of 1/b+10
my bad \[\frac{ -7 }{ b+10 }\]
Yep. Correct! Good job.
Ha, there you go eb :)
thanks
No worries.
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