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#1 2007-02-07 15:17:26

AlgebraIIisHard
Member
Registered: 2007-02-07
Posts: 1

Factoring (AlgebraII)

Can anyone make factoring easy for me? I'm having a lot of trouble with it. hmm

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#2 2007-02-07 16:43:57

Jai Ganesh
Administrator
Registered: 2005-06-28
Posts: 45,956

Re: Factoring (AlgebraII)

I guess you're referring to prime factorisation.
Prime numbers are numbers which are divisble only by 1 and themselves. The other numbers are called non-primes or composite numbers.
Example:-
Prime factorisation of 42:-
Try dividing it by 2. You get 21 as the quotient.
Therefore, 42=21x2.
Try dividing 21 by 3 (since it is an odd number, it isn't divisible by 2).
You get 21=3x7
Since 7 is also a prime number,
42 can be factorised as
42=2x3x7.
Similarly, 24 can be factorised as
24=2x2x2x3
If you have any specific queries, you may post here again.


It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#3 2007-02-07 17:47:39

justlookingforthemoment
Moderator
Registered: 2005-05-26
Posts: 2,161

Re: Factoring (AlgebraII)

Or maybe you mean factoring, as in 'factorise 12x² + 4x + 3'?

There are many different techniques which help factorise different problems, so it might be more helpful if you tell us what you need help factoring.

But a general, starting rule is look for a common factor.

Eg. Factorise 2x + 6.
First, we would check if our two terms, 2x and 6, have a common factor. And they do. 2x and 6 can both be divided by 2, which makes 2 a common factor. So 2 can be taken out by dividing each term by two, which gives us 2(x + 3).

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#4 2007-02-17 04:20:10

dmatthias
Member
Registered: 2007-02-17
Posts: 4

Re: Factoring (AlgebraII)

Can anyone make factoring easy for me? I'm having a lot of trouble with it.

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#5 2007-02-17 04:21:10

dmatthias
Member
Registered: 2007-02-17
Posts: 4

Re: Factoring (AlgebraII)

18x+21x+6

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#6 2007-02-17 04:23:18

dmatthias
Member
Registered: 2007-02-17
Posts: 4

Re: Factoring (AlgebraII)

dmatthias wrote:

18xsquared+21x+6

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#7 2007-02-17 04:25:14

dmatthias
Member
Registered: 2007-02-17
Posts: 4

Re: Factoring (AlgebraII)

Can anyone make factoring easy for me? I'm having a lot of trouble with 18x+21x+6

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#8 2007-02-17 06:26:25

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Factoring (AlgebraII)

When you registered to this forum, this would have been the first thing you saw:

This is a Forum, it is not instant chat. Leave your message, and come back later (hours or days) to see what responses you got.

You're expecting an answer far too quickly, and just because you don't get one that doesn't mean that you can abuse the report button.

Anyway, here's your question.
18x² + 21x + 6.

We can make the factorising a bit easier by taking out the common constant of 3:
3(6x² + 7x + 2).

Now, to factorise the bracket we need to find two numbers that add to give 7 and multiply to give 12 (because 6x2 = 12).
These numbers are 4 and 3.

Separating the 7x using these gives us (6x² + 4x) + (3x + 2).

The two terms in the brackets both have are both divisible by 3x+2, so we can factorise that out:
(3x+2)(2x) + (3x+2)(1) = (3x+2)(2x+1)

And there we go. Successful factorisation. Now we just need to put the 3 back and we'll be done:
3(3x+2)(2x+1)

Good things come to those who wait. smile


Why did the vector cross the road?
It wanted to be normal.

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