anyone any good at algebra?

cm1179

cm1179

cm1179

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Joined
30 Sep 2009
Posts
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anyone give me any pointers on any basics to remember? i just can't get my head around it. we've got to do quadractic equations and factorisation etc on my college course and i just can't get to grips with it
 
Factorisation is easy. I think! Give me an equation and I will see if I know how to do it.
 
thanks. i did some searches and got reaults such as elephant maths, bit of a wake up call as i'm 30 and did my gcse's in 1996. me and algebra never did get on
 
cm1179 said:
(that's 3p squared, can't do the small 2)
Click to expand...

simples:

ALT+0178 on the num pad

like this: x² :D

oh and cubed is ALT+0179 if you care ³ :p


3p² + p - 2 = (3p-2)(p+1)

Steps:
  • "3p²" can only be generated by multiplying "p" and "3p" together (there are no other integer solutions given three is prime), therfore your brackets must be (3p+a)(p+b)
  • "-2" can be generated by multiplying -1 and 2 together or -2 and 1. therefore one set of these are your a and b above.
  • given you have (3p*b)+(p*a) to generate the middle term "+p" the only solution is a=+1 b=-2 as this gives (3p)+(-2p) = +p
 
Last edited:
cm1179 said:
anyone give me any pointers on any basics to remember? i just can't get my head around it. we've got to do quadractic equations and factorisation etc on my college course and i just can't get to grips with it
Click to expand...

Start off with the simple ones. My example it's

x^2+7x+12


So you know you're going to have two sets of brackets:
()() <- ha ha looks like boobs:p

Okay so there's no integer in front of the x^2 to it's x's in both
(x ) (x ) <- this is getting ruder

Now the two numbers in here must MULTIPLY to get 12 and must add to get 7

So what numbers satisfy that?
4 and 3

Hence the factorization of x^2+7x+12

is (x+4)(x+3)
 
dangerstat said:
Start off with the simple ones. My example it's

x^2+7x+12


So you know you're going to have two sets of brackets:
()() <- ha ha looks like boobs:p

Okay so there's no integer in front of the x^2 to it's x's in both
(x ) (x ) <- this is getting ruder

Now the two numbers in here must MULTIPLY to get 12 and must add to get 7

So what numbers satisfy that?
4 and 3

Hence the factorization of x^2+7x+12

is (x+4)(x+3)
Click to expand...

so i only need to concentrate on the figures to the far right, in this case, the 7 and 12? (it's things like this which are going straight over my head ) and is it always the number on the end that needs to be obntained by multiplication?
 
dangerstat said:
Start off with the simple ones. My example it's

x^2+7x+12


So you know you're going to have two sets of brackets:
()() <- ha ha looks like boobs:p

Okay so there's no integer in front of the x^2 to it's x's in both
(x ) (x ) <- this is getting ruder

Now the two numbers in here must MULTIPLY to get 12 and must add to get 7

So what numbers satisfy that?
4 and 3

Hence the factorization of x^2+7x+12

is (x+4)(x+3)
Click to expand...

Good explanation! I think I'll leave it to you to teach him xD.
 
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