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maths question - factorising


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#1 Quick hedgehog

Posted 25 August 2019 - 02:07 PM

DH has been helping DS (year 8) with some maths homework that involves factorising and they are stuck on the last question.

The question requires the following to be factorised in groups

bc - 1 + c - b

We know the answer (b+1) (c-1) but can't figure out how to get there.

Is anyone able to supply the working out so we can see how it is done?

Thanks

#2 ~Kay~

Posted 25 August 2019 - 02:23 PM

bc + c - b - 1 = c (b +1) -1 (b +1) = (b+1)(c-1)

#3 Quick hedgehog

Posted 25 August 2019 - 02:34 PM

Thanks Kay.

DS has given up entirely, DH is still scratching his head over it and trying to work out why it is so.   :)

#4 Riotproof

Posted 25 August 2019 - 02:39 PM

My working had one extra line that might help.

Bc - 1 + c -b = c (b + 1) -1(b + 1)  

Then take the c-1 and multiply by the b+1 to get your answer

= (C-1)(b+1)

#5 mumbag

Posted 25 August 2019 - 02:55 PM

Here's my attempt at explanation, starting with:

bc - 1 + c - b

bc and c have a common factor c so can be written as c(b+1)

1 and b have a common factor 1 so can be written as 1(b+1)

so:

c(b+1)-1(b+1)

So now you have a common factor (b+1) so you can rewrite as:

(c-1)(b+1)

Edited by mumbag, 25 August 2019 - 02:55 PM.


#6 Quick hedgehog

Posted 25 August 2019 - 03:22 PM

what happens if using BC and B with B as the common factor.  What I am struggling with is the use of the addition and subtraction symbols and which one to use where.  I can't seem to get this version to work so don't understand why we chose the common factor sequences we do.

#7 mumbag

Posted 25 August 2019 - 03:34 PM

I'm honestly not an expert, just a fellow learner, so take this with a grain of salt!

I think maybe it would make more sense if I had written 1(b+1) as -1(b+1) in the explanation above.

The terms you need to be equivalent to are:

+bc
-1
+c
-b

So:

c(b+1) = +bc +c
-1(b+1) = -b -1

but

b(c+1) = +bc +b
-1(c+1) = -c -1

So +c and -b are coming out as -c and +b if you use "BC and B with B as the common factor".

Does that make sense?

Edited by mumbag, 25 August 2019 - 03:43 PM.


#8 mumbag

Posted 25 August 2019 - 03:50 PM

Actually, I kept looking at it and you can use bc and b with b as the common factor and get the same answer.

b(c-1) = +bc - b
+1(c-1) =+c -1

=b(c-1) + 1(c-1)

=(b+1)(c-1)

Edited by mumbag, 25 August 2019 - 03:50 PM.


#9 Quick hedgehog

Posted 25 August 2019 - 08:22 PM

Thanks mumbag.  DH is out now, I'll show him tomorrow. He doesn't like to be beated by maths .  :)

#10 Sincerely

Posted 25 August 2019 - 09:35 PM

View PostQuick hedgehog, on 25 August 2019 - 03:22 PM, said:

what happens if using BC and B with B as the common factor.  What I am struggling with is the use of the addition and subtraction symbols and which one to use where.  I can't seem to get this version to work so don't understand why we chose the common factor sequences we do.

This post is a bit long but outlines the steps one by one: Starting at basics with numbers instead of letter variables:

2(3-1) is shorthand for 2x(3-1) which can be solved two ways:

You can work out what's in the brackets first:
2x(3-1) = 2x2 = 4

Or you can expand before doing the subtraction:
2x(3-1) = 2x3 - 2x1 = 6-2 = 4

Similarly, if we use letter variables instead of numbers:
bx(c-1) = bxc - bx1

Or in shorthand:
b(c-1) = bc-b

Therefore, in reverse, we can take out the common factor of b:
bc-b = b(c-1)
_________________________________________
The original problem was bc-1+c-b
Rearrange this to: bc-b+c-1

=b(c-1)+c-1 = b(c-1) +(c-1) = bx(c-1) + 1x(c-1)

Note: Adding brackets & multiplying by 1 changes nothing.
________________________________________

Now, if we again looked at numbers instead of letters:
5x(3-1)+1x(3-1) = 5x2 + 1x2 = 10+2 = 12

The other way to solve by taking out the common factor of 2:
5x(3-1)+1x(3-1) = 5x2 + 1x2 = (5+1)x2 = 6x2 = 12

So, we can see that: 5x(3-1)+1x(3-1) = (5+1)x(3-1)

Similarly, bx(c-1) + 1x(c-1) = (b+1)x(c-1)


#11 Sincerely

Posted 26 August 2019 - 07:35 PM

Is your DS satisfied that he now understands the general concepts? If not, perhaps you could post other questions & by working through a few examples, it might become clear.




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