Thread: Factorising mistake in Edexcel higher plus book?

1. Factorising mistake in Edexcel higher plus book?

Well exams in 2 days and i forgot how to factorise. I decided to do some examples from da book and i saw this one
Factorise
4(p-q) + (p-q)^3 (^ = to the power btw:P)
I checked the back and it told me that the answer is (p-q)[4+(p-q)^2] That makes no sense whatsoever anyway we would left with pq^2 and qp^2 ffs i need an A badly and this question is making me nuts help ?!

2. Hello,

Originally Posted by Mobix
Well exams in 2 days and i forgot how to factorise. I decided to do some examples from da book and i saw this one
Factorise
4(p-q) + (p-q)^3 (^ = to the power btw:P)
I checked the back and it told me that the answer is (p-q)[4+(p-q)^2] That makes no sense whatsoever anyway we would left with pq^2 and qp^2 ffs i need an A badly and this question is making me nuts help ?!
$4(p-q) + (p-q)^3=4{\color{red}(p-q)}+{\color{red}(p-q)}(p-q)^2={\color{red}(p-q)}[4+(p-q)^2]$

isn't it ?

3. thanks for help that makes still no sense i wasnt in when we did that in school .
Could you tell me whats the function of a square bracket in factorising please?
wait wait 4(p-q)+ (p-q)(p-q)2 if we are multiplying than it should be 4(p-q) + p3-q3 + we would still have pq3 qp3 ??

4. AHHH but when that wouldnt be factorysing that would be kind of expanding wouldnt it?

5. Originally Posted by Mobix
thanks for help that makes still no sense i wasnt in when we did that in school .
Could you tell me whats the function of a square bracket in factorising please?
...
Factorizing is the contrary of expanding (and vice versa )

You extract a common factor of a sum (or a product as well).

If you have chosen for the common factor (or you are told what the common factor has to be) then you must divide each summand by the factor and the result of the division is written into the bracket.

Example:

$21a^2b+35ab^2-42a^2b^2$

I take as common factor $7ab$

$21a^2b+35ab^2-42a^2b^2 = 7ab \left(\frac{21a^2b}{7ab} +\frac{35ab^2}{7ab} - \frac{42a^2b^2}{7ab} \right)$

Simplify the fractions and you'll get the final result:

$21a^2b+35ab^2-42a^2b^2 = 7ab(3a+5b-6ab)$

You even can factor out a factor which doesn't appear in the sum:

Factor out x of

$a+b+c = x\left(\frac ax+\frac bx + \frac cx\right)$

Maybe this helps you a little bit(?)