Question

can anyone explain some super factorisation?

k(2k-1)(2k+1)+3(2k+1)
all on 3

moves to

(2k+1)(2k²-k+6k+3)
all on 3

moves to

(2k+1)(2k+3)(k+1)
all on 3

moves to

(k+1)(2k²+5k+3)(k+1)
all on(divide by)3

then that goes to

(K+1)(4(k+1)²-1)
all on 3

Answer 1

please dont explain it all at once...

Answer 2

not sure what you're trying to do...

Answer 3

im just trying to understand each of those steps

Answer 4

ok first step: both terms have a (2k+1) so it can be factored out or divide each term by it
-> (2k+1)(k(2k-1) +3)
Then they distribute the k(2k-1)
->(2k+1)(2k^2 -k +3)
**Not sure where they got the 6k from**

Answer 5

its when you * 3(2k+1)^2

cause you are adding the fractions, you had to make them have the same denominator ... isnt it?

Answer 6

oh its squared?
no you dont have to worry about adding fraction, i believe its one fraction already with 3 as denominator

Answer 7

yeah sorry.

Answer 8

it is squared, and yeah that will clear up my mistaken fraction problem too.

Answer 9

ok then after dividing out the (2k+1)
-> (2k+1)(k(2k-1) + 3(2k+1))
Distribute
->(2k+1)(2k^2 -k +6k +3)
Add like terms
->(2k+1)(2k^2 + 5k +3)

Answer 10

Next step:
Factor (2k^2 + 5k +3)
2*3 = 6
2+3 = 5
Rewrite 5k as 2k+3k
2k^2 +2k +3k +3
2k(k+1) +3(k+1)
(2k+3)(k+1)
this gives
-> (2k+1)(2k+3)(k+1)

Answer 11

oh

Answer 12

thats powerful isnt it.

Answer 13

from there im not sure where they're going
at this point its completely factored

Answer 14

well the whole line is
(k+1)(2k+1)(2k+3)(k+1)

so that can be (factorised? is that the word?)
into
(k+1)(4(k+1)²-1)

all/3

Answer 15

it is completely factored, you're right, but this isn't the whole question (its most of a mathematical induction question so i have to get the above equation to equal another equation... thats why they went further.

Answer 16

oh ok i get it now
that next line threw me off because it had an extra k
Next step:
multiply out (2k+1)(2k+3) = 4k^2 +8k +3
then use completing the square
factor out a 4 from first 2 terms
->4(k^2 +2k) + 3
take 2nd coefficient, halve it then square it
->4(k^2 +2k + 1) + 3 - 4
->4(k+1)^2 -1

Answer 17

oh btw, i believe its "factored" not "factorised"

Answer 18

thats alright, i complain when people say math instead of maths

but what pray tell, is completing the square?

Answer 19

its a method for solving a quadratic equation, also it changes the form from
ax^2 +bx +c -->a(x-h)^2 + k
it uses the idea of a perfect square: (x+c)^2 = x^2 +2cx + c^2
notice the middle coefficient of 2c
if i take half of 2c i get c, if i then square it i get c^2 which is the constant term

Answer 20

can you please give one example, not this current one?

Answer 21

ok
x^2 + 6x - 2
middle coefficient is 6
take half and square it....6/2 = 3 .....3^2 = 9
so add 9, however we also subtract 9 because you dont want the value the equation to change
(x^2 + 6x + 9) -2 -9
the part in parenthesis can be written as a square, just use half of 6
(x+3)^3 - 11
This is equivalent to what we started with, its just in a different form

Answer 22

oops that should be "^2" at the end

Answer 23

oh good, i was looking twice at that bit.

Answer 24

another thing, completing the square only works if its "x^2" not "3x^2"
If there is a number in front of the x^2 you have to factor it out first then do the steps

Answer 25

oh ok

Answer 26

any other questions

Answer 27

why didnt i learn this ten years ago

Answer 28

(no)

Answer 29

haha i dunno...it should be standard in Algebra 2

Answer 30

i think that might be a problem with aussie maths classes (we just have everything lumped together and its up to teh teacher to decide what is appropriate to learn, from grade 1-12 and then you're off to uni to find out how much you should know/already know)

Answer 31

oh i see, im from the USA and education is all about standards right now which helps keep teachers accountable kinda

Answer 32

anyway, i can complain about that later, i think i am set apart from the norm of what most other australians know about maths.

Answer 33

alright good luck to you

Answer 34

thanks

Answer 35

again, you probably deserve like 10 medals for answering my question
haha