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How to find factors of big numbers quickly

Like finding factors of 140 which add up 4.

Just doing past papers and came across this and wondering if this actually happens I would spend so long just finding factors
(edited 12 years ago)
Reply 1
Avatar for praj1
praj1
1
that makes no sense lol?

why the neg rep?.....it dont make sense?.....if i find any 2 factors of 140...they'll never add up to 4 ?....i was just asking so i could actually attempt the question
(edited 12 years ago)
Reply 2
Avatar for hayzelle
hayzelle
9
You could split it into its prime factors and multiply them together in different ways to find factors, that's what I'd probably do.

Though if you have a calculator for a number like 140 it shouldn't be that hard: clearly it divides by 10 so 10,14 are factors. You can find between 10 and 14 there are no factors, so just divide 140 by numbers less than 10 to get 7+20, 5+28, 4+35, 2+70, and of course 1+140.

I don't really understand the 'add up to 4' part though?
Reply 3
Avatar for Skaterkid
Skaterkid
OP
9
Original post by hayzelle
You could split it into its prime factors and multiply them together in different ways to find factors, that's what I'd probably do.

Though if you have a calculator for a number like 140 it shouldn't be that hard: clearly it divides by 10 so 10,14 are factors. You can find between 10 and 14 there are no factors, so just divide 140 by numbers less than 10 to get 7+20, 5+28, 4+35, 2+70, and of course 1+140.

I don't really understand the 'add up to 4' part though?


Part of an equation. You need to find factors of 140 which add up 4 inorder to factorise it
Reply 4
Avatar for hayzelle
hayzelle
9
Original post by Skaterkid
Part of an equation. You need to find factors of 140 which add up 4 inorder to factorise it


Could you post the equation? Still a bit confused, sorry.
Reply 5
Avatar for Skaterkid
Skaterkid
OP
9
Original post by hayzelle
Could you post the equation? Still a bit confused, sorry.


5y^2+4y-28=0
Original post by Skaterkid

Original post by Skaterkid
5y^2+4y-28=0


Are you sure you have to factorise it?
Original post by Skaterkid
5y^2+4y-28=0


Well 5 is prime so you need 5y in one bracket and y in the other. The only way you could get -28 at the end is to multiply a positive by a negative so you know the signs you need. Now all you have to do is consider factors of 28 and there aren't that many of them!
Reply 8
Avatar for hayzelle
hayzelle
9
5y2+4y−28=05y^2+4y-28=0

You're making it very hard for yourself!

Instead of looking for factors of 140, concentrate on the coefficient of y^2 and the constant separately.

So the factors of 5 are only 1 and 5, so you must have: (5y+a)(y+b)
The constant term is a*b so look for a and b that multiply to make -28 and also give +4y in the middle.

If you're really struggling with factorising a particular equation remember you can always use the quadratic equation.
Reply 9
Avatar for steve2005
steve2005
17
Original post by Skaterkid
Like finding factors of 140 which add up 4.

Just doing past papers and came across this and wondering if this actually happens I would spend so long just finding factors


Having glanced at this thread. You should have said factors of minus 140 that sum to 4. ( 14 and minus 10)

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