Year 9 simplification maths

K.I.L.E.R

Retarded moron
Veteran
Which I haven't touched in a long time(6 years) but I now need it as I Tuesday's test will have questions on induction and that requires simplification of basic algebra.

(1/3)k(k+1)(K+2) + (k+1)(K+2)

Can someone show me each step of simplification of this.
Thanks.
 
I thought of factorising, not just multiplying it all the way through. Having squares and not eliminating the addition is not simplifying to me :???:
 
Sorry but it has been 6 years since I touched this stuff.
I've practically forgotten majority of this stuff.

So if I had:

6/4 * ba * (ba + 1)

it can be turned into:

( (6/4)ba + 1 )

?
 
K.I.L.E.R said:
Which I haven't touched in a long time(6 years) but I now need it as I Tuesday's test will have questions on induction and that requires simplification of basic algebra.

(1/3)k(k+1)(K+2) + (k+1)(K+2)

Can someone show me each step of simplification of this.
Thanks.

Hmm, it kind of depends what you consider simpler. Do you need to factorise it, or multiply it out to produce a cubic equation.
If you are multiplying it out then the first thing to do is calculate (k+1)(K+2)
which should be kk + 3k +2

so replacing that we have
k/3(kk+3k+2) + (kk+3k+2)

=1/3( k(kk+3k+2) + 3(kk+3k+2))
=1/3( kkk+3kk+2k +3kk +9k+6)
=1/3(kkk+6kk+11k+6)

Alternatively if you are wanting to factorise it:

Well the (K+1)(K+2) is a common factor so you could take that out.
so that would be (k/3 +1)(K+1)(K+2)


Note it is about 10 years since I have done this kind of maths so chances are I have made a mistake :)
 
K.I.L.E.R said:
Sorry but it has been 6 years since I touched this stuff.
I've practically forgotten majority of this stuff.

So if I had:

6/4 * ba * (ba + 1)

it can be turned into:

( (6/4)ba + 1 )

?

Not really. Something missing from the first expression, which is by the way already the factorised form.
 
capt chickenpants has it i think (at first glance). depending on what simplified means to europeans (i know, austrailia is it's own continent, still you used to be european convicts, and in us we don't have year 9 math)....reduction of terms or factoring or...forgetting?


*edit*

uhm

so replacing that we have
k/3(kk+3k+2) + (kk+3k+2)

=1/3( k(kk+3k+2) + 3(kk+3k+2))

doesn't the 1/3 go away when you multiply the whole line by 3 (simplifier). so you end up with just k**3 + 6k**2 + 11k + 6?
 
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AFAIK algebraic simplification isn't the same as factorisation. Though its one of those maths definition that I don't like so much.

Cartoon Corpse said:
k/3(kk+3k+2) + (kk+3k+2)

= 1/3( k(kk+3k+2) + 3(kk+3k+2))

That doesn't equal. How did you end up with 3(kk+3k+2) ?

doesn't the 1/3 go away when you multiply the whole line by 3 (simplifier). so you end up with just k**3 + 6k**2 + 11k + 6?

You can only do that if the original question is (1/3)k(k+1)(K+2) + (k+1)(K+2) = 0, but it isn't so you can't.

I'll do it for Killer sake. Here is my version of it, take it as you will. Simplification in algebra is collecting like terms, its different from factorising. So if you are given something in factorised form, expand it out and afterward put the like terms together by addition or subtraction.

(1/3)k(k+1)(k+2) + (k+1)(k+2)
= (1/3)k(k^2 + 2k + k + 2) + (k^2 + 2k + k + 2)
= (1/3)k(k^2 + 3k + 2) + (k^2 + 3k + 2)
= (1/3)(k^3 + 3k^2 + 2k) + (k^2 + 3k + 2)
= 1/3k^3 + k^2 + 2/3k + k^2 + 9/3k + 2 (edited)
= 1/3k^3 + 2k^2 + 11/3k + 2 (edited)

There you go.

Edit: Sorry I made a typo on the second last line that lead to an error in the final line.
 
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Captain Chickenpants said:
Note it is about 10 years since I have done this kind of maths so chances are I have made a mistake :)
Hey! If you want some practice, pop up to our office - I've got some crackers of equations that need simplifying :)
 
Thanks guys.

When I'm asked to simplify something and I can do it by factorisation or algebraic simplification, does it matter which one I do?
 
K.I.L.E.R said:
Thanks guys.

When I'm asked to simplify something and I can do it by factorisation or algebraic simplification, does it matter which one I do?

Maths is very formal, so it does matter what you perform. If you're asked to perform simplification, you do simplification, if you're asked to perform factorisation, you do factorisation. If they tell you, you can choose, choose one. I would go with simplification since its easier, but I find factorised form is more useful.
 
It's just semantics. As such, both are simplification of a kind. Factorisation is more useful in the computer world, makes algorithms simpler and faster by reducing higher-order calculations. For "normal" maths, simplification by multiplying it all through etc. is more useful mostly, since it makes calculating easier for humans. For the machines it's rather the opposite.
 
V3 said:
That doesn't equal. How did you end up with 3(kk+3k+2) ?
It does. Note the parenthesis surroundnig everything but 1/3.
That way of doing it makes it a bit easier since he doesn't have to care about fractions. Maybe it was because of that he dodged the error you made in the last step (11/3 k vs 5/3 k).

About the meaning of "simplification":
I can't say for sure about the English term, but the Swedish term that I thought was equivalent, is actually subjective. It's about making things look neat. Expanding polynoms and collecting terms is just one way of doing it.
 
Basic said:
About the meaning of "simplification":
I can't say for sure about the English term, but the Swedish term that I thought was equivalent, is actually subjective. It's about making things look neat.

Agreed. Mathsworld, for example, has a text where "simplification" means subsitituting new symbols for whole sub-expressions.
 
Basic said:
It does. Note the parenthesis surroundnig everything but 1/3.
That way of doing it makes it a bit easier since he doesn't have to care about fractions. Maybe it was because of that he dodged the error you made in the last step (11/3 k vs 5/3 k).

Ahh yes, I missed the bracket and made mistake in my example. Not use to working with this notation.

About the meaning of "simplification":
I can't say for sure about the English term, but the Swedish term that I thought was equivalent, is actually subjective. It's about making things look neat. Expanding polynoms and collecting terms is just one way of doing it.

Yeah, this is something about maths that I don't like.
 
V3 said:
AFAIK algebraic simplification isn't the same as factorisation. Though its one of those maths definition that I don't like so much.



That doesn't equal. How did you end up with 3(kk+3k+2) ?


You can only do that if the original question is (1/3)k(k+1)(K+2) + (k+1)(K+2) = 0, but it isn't so you can't.

I'll do it for Killer sake. Here is my version of it, take it as you will. Simplification in algebra is collecting like terms, its different from factorising. So if you are given something in factorised form, expand it out and afterward put the like terms together by addition or subtraction.

(1/3)k(k+1)(k+2) + (k+1)(k+2)
= (1/3)k(k^2 + 2k + k + 2) + (k^2 + 2k + k + 2)
= (1/3)k(k^2 + 3k + 2) + (k^2 + 3k + 2)
= (1/3)(k^3 + 3k^2 + 2k) + (k^2 + 3k + 2)
= 1/3k^3 + k^2 + 2/3k + k^2 + 9/3k + 2 (edited)
= 1/3k^3 + 2k^2 + 11/3k + 2 (edited)

There you go.

Edit: Sorry I made a typo on the second last line that lead to an error in the final line.

That doesn't equal. How did you end up with 3(kk+3k+2) ?

i believe i was just questioning what chickenpants had written. showing that multiplying through by 3 would get rid of the 1/3, not leave it.


the original equation posted was:
(1/3)k(k+1)(K+2) + (k+1)(K+2)

NOT (1/3)k(k+1)(K+2) + (k+1)(K+2) = 0

so, given that 3* the entire equation still preserves the algebraic relationship and allows for further combination of terms, just alters the magnitude. if there is no established equivalence, i think it's only the empirical relationship (algegra) that is in need of preservation.

DOH i just noticed there are lower case AND upper case 'k's. is this a trick?
 
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God, people.

Long time.

(1/3)k(k+1)(k+2) + (k+1)(k+2)=
(k/3+1)(k+1)(k+2)= ---> see Captain Chickenpants
(k/3+3/3)(k+1)(k+2)=
1/3(k+3)(k+1)(k+2)= ---> reorder to look nicer
(k+1)(k+2)(k+3)/3

which will be an integer as long as k is an integer

Enjoy,

V
 
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vb said:
God, people.

Long time.

(1/3)k(k+1)(k+2) + (k+1)(k+2)=
(k/3+1)(k+1)(k+2)= ---> see Captain Chickenpants
(k/3+3/3)(k+1)(k+2)=
1/3(k+3)(k+1)(k+2)= ---> reorder to look nicer
(k+1)(k+2)(k+3)/3

which will be an integer as long as k is an integer

Enjoy,

V

Yup, you're the man! :smile:
 
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