Expansion Identities... (1 Viewer)

[Lukybear]

I was just doing a few Qs from Cambridge, Chapter 1 and i found this question, which really bothered me for a while.

x^6-64

After a few hours slaving away at this question, i finally got it. But it required knowing the identity of

x^4+a^2.x^2+b^2 = (x^2-2x+4)(x^2+2x+4)

Because i havent been taught this and its not in the book + i couldnt see any connection to normal quadratics i thought mabey Cambridge is overteaching. On the other hand i could just be very very stupid. Is this something that i should know for yr 12?

Also just curious, how would you guys do that question?

 

[Iruka]

64=4^3

Its the difference of two cubes in disguise.

 

[Lukybear]

64=4^3

Its the difference of two cubes in disguise.

Yea, thats part of the Question, but it was factorise it fully...

ANd i got

x^6-64
(x^2-4)(x^4+4x^2+16)
=
(x-2)(x+2)(x^2+2x+4)(x^2-2x+4)
QED

The bit in bold is where i got confused...

 

[lyounamu]

I was just doing a few Qs from Cambridge, Chapter 1 and i found this question, which really bothered me for a while.

x^6-64

After a few hours slaving away at this question, i finally got it. But it required knowing the identity of

x^4+a^2.x^2+b^2 = (x^2-2x+4)(x^2+2x+4)

Because i havent been taught this and its not in the book + i couldnt see any connection to normal quadratics i thought mabey Cambridge is overteaching. On the other hand i could just be very very stupid. Is this something that i should know for yr 12?

Also just curious, how would you guys do that question?

x^6 - 64 = (x^2)^3 - 4^3 = (x^2-4)(x^4+4x^2+16)
= (x^2 - 4) (x^4 + 8x^2 - 4x^2 + 16)
= (x^2-4) ((x^2+4)^2 - (2x)^2)
= (x^2-4)(x^2+4+2x)(x^2+4-2x)


As Iruka already mentioned, this is a difference between two cubes. And the subsequent working follows. (even though I don't see much point).

 

[Lukybear]

x^6 - 64 = (x^2)^3 - 4^3 = (x^2-4)(x^4+4x^2+16)

As Iruka already mentioned, this is a difference between two cubes.

Yea but thats not factorised fully, as mentioned in the above post... sorry 2 lazy to type

 

[lyounamu]

Yea but thats not factorised fully, as mentioned in the above post... sorry 2 lazy to type

I actually finished that above. (had to re-touch it because I didn't see your last point. BUT I don't see the point in further factorising that.

EDIT: (x^2-4)(x^4+4x^2+16) = (x^2-4)(x^4 + 4x^2 +16 ) = (x+2)(x-2)(x^2+4+2x)(x^2+4-2x)

THIS IS THE fully factorised version.

 

[Lukybear]

x^6 - 64 = (x^2)^3 - 4^3 = (x^2-4)(x^4+4x^2+16)
= (x^2 - 4) (x^4 + 8x^2 - 4x^2 + 16)
= (x^2-4) ((x^2+4)^2 - (2x)^2)
= (x^2-4)(x^2+4+2x)(x^2+4-2x)

wow... can see why u can first in state for 2 unit... 2 genious namu...

but i got 1 more Q, do you learn that or did you firgure it out all by urself?

 

[lyounamu]

wow... can see why u can first in state for 2 unit... 2 genious namu...

but i got 1 more Q, do you learn that or did you firgure it out all by urself?

lol,

I didn't learn it. You don't really learn it. You just gotta subtract or add certain parts to the equation so that you can find the way to factorise it.

So here, I just came up with 8x^2 - 4x^2 instead of 4x^2 so that I can actually factorise it.

But I don't think it would be necessary...unless you do 4 unit question where they actually ask you to find the individual roots.

 

[Lukybear]

Still kinda brilliant...

but i also have one more question, how did

(x^2+4)^2-(2x)^2

become
(x^2+4+2x)(x^2+4-2x)

o wait nvm... is it because -(2x)^2 = -2x.2x

 

[Iruka]

OK, I see what you mean. I didn't actually do the question.

As a consequence of the Fundamental Theorem of Algebra and the fact that polynomials with real coefficients have roots that come in complex conjugate pairs, any polynomial with real coefficients can be (theoretically, at least) factored over the reals into linear and quadratic factors. But I hardly think that you are expected to know this in 2U! I guess Namu realized that you could factorise the quartic further and used a bit of ingenuity to work out how to actually do it...

Cambridge books do contain a lot of material that goes beyond the syllabus, particularly in the challenge questions.

Practice for Lukybear (not Namu, - he will probably get it in about 2 seconds):

Factorise x^4+1.

 

[lyounamu]

Still kinda brilliant...

but i also have one more question, how did

(x^2+4)^2-(2x)^2

become
(x^2+4+2x)(x^2+4-2x)

o wait nvm... is it because -(2x)^2 = -2x.2x

Difference of two squares

x^2 - y^2 = (x+y)(x-y)

similarly:

(x+2)^2 - y^2 = (x+2+y)(x+2-y)

 

[lolman12567]

= (x³)² - (8)²
= (x³ - 8)(x³ + 8)
= (x - 2)(x² + 2x + 4)(x + 2)(x² - 2x + 4)

 

[lolokay]

difference of 2 squares, then difference and sum of 2 cubes.. always do the difference of 2 squares before the 2 cubes, or it will be much harder

 

[addikaye03]

Factorise x^4+1
x^4-i^2
(x^2-i)(x^2+i)
[(x-rti)(x+rti)](x^2+i)

..was the Q supposed to be x^4-1, or is there a two unit method to doing this Q

IF, it was x^4-1
(x^2)^2-(1^2)^2
(x^2-1)(x^2+1)
(x-1)(x+1)(x^2+1)... i still doubt that would be asked in 2U

 

[kaz1]

Factorise x^4+1
x^4-i^2
(x^2-i)(x^2+i)
[(x-rti)(x+rti)](x^2+i)

..was the Q supposed to be x^4-1, or is there a two unit method to doing this Q

IF, it was x^4-1
(x^2)^2-(1^2)^2
(x^2-1)(x^2+1)
(x-1)(x+1)(x^2+1)... i still doubt that would be asked in 2U

2unit does not deal with complex numbers.

x⁴+1
x⁴+2x²+1-2x²
(x²+1)²-2x²
(x²+rt2x+1)(x²-rt2x+1)

 

[addikaye03]

2unit does not deal with complex numbers.

x⁴+1
x⁴+2x²+1-2x²
(x²+1)²-2x²
(x²+rt2x+1)(x²-rt2x+1)

yeah i knew that, thats why i figured there must have been a 2unit method.. i didnt see that way.. i do now..nice work

 

[Lukybear]

Wow, nice work guys, much more brilliant then me, all of you people.

But considering my intelligence, dont think thats a compliment.

Guess more practise is needed.

 

[lyounamu]

Hm...I didn't know that question asked by Iruka was for us...(it was intended for luckybear).
And yeah, kaz1's solution is more technically correct in this context (as this forum is a 2 unit forum and also considering the OP's current level of maths where he is currently involved in 2/3 units of maths)

 

[Lukybear]

Well these expansion techniques be required for the 4 Unit math??

 

[lyounamu]

Well these expansion techniques be required for the 4 Unit math??

Yeah. But you will learn all these or experience all these stuff before you learn 4 Unit anyway.