factorising help (1 Viewer)

_kuroo · Mar 6, 2012

ok i dont have 1 question, but MULTIPLE questions on factorising. im really lucky that my test has been postponed till thursday. i dont know how to do it so putting up some rules (like x^a*x^b=x^a+b) to follow and tips and all the steps inbetween would be nice

i) x^3-9xy^2z^2

ii) 2ap^3 - 2aq^3

iii) 27a^6-28a^3+1

nightweaver066 · Mar 6, 2012

$i) x^3 - 9xy^2z^2$

$It doesn't look factorisable using any fancy methods like difference of two squares or two cubes or anything of the sort right now, so lets first take out a common factor between them two; x$

$x(x^2 - 9y^2z^2)$

$ooh. a difference of two squares.$

$x(x - 3yz)(x + 3yz)$

$ii)2ap^3 - 2aq^3$

$Hey, this looks like a difference of two cubes! Let's take out a factor first$

$2a(p^3 - q^3)$

$Difference of two cubes. :)$

$2a(p - q)(p^2 + pq + q^2)$

$iii) 27a^6 - 28a^3 + 1$

$This somewhat resembles a quadratic to me. Let's try letting u = a^3.$

$$Remember that $ a^6 = (a^3)^2$

$27u^2 - 28u + 1$

$Nice quadratic. Let's factorise..$

$27u^2 - 27u - u + 1$

$(27u - 1)(u - 1)$

$Subbing back in u = a^3$

$(27a^3 - 1)(a^3 - 1)$

$oh hey, more difference of two cubes!$

$(3a - 1)(9a^2 + 3a + 1)(a - 1)(a^2 + a + 1)$

$And of course rearranging to make it look presentable$

$(a -1)(3a - 1)(a^2 + a + 1)(9a^2 + 3a + 1)$

_kuroo · Mar 6, 2012

wow thanks. btw how do you make your text look like that?(where the powers dont have arrows and stuff)

nightweaver066 said:
$i) x^3 - 9xy^2z^2$

$It doesn't look factorisable using any fancy methods like difference of two squares or two cubes or anything of the sort right now, so lets first take out a common factor between them two; x$

$x(x^2 - 9y^2z^2)$

$ooh. a difference of two squares.$

$x(x - 3yz)(x + 3yz)$

$ii)2ap^3 - 2aq^3$

$Hey, this looks like a difference of two cubes! Let's take out a factor first$

$2a(p^3 - q^3)$

$Difference of two cubes. :)$

$2a(p - q)(p^2 + pq + q^2)$

$iii) 27a^6 - 28a^3 + 1$

$This somewhat resembles a quadratic to me. Let's try letting u = a^3.$

$$Remember that $ a^6 = (a^3)^2$

$27u^2 - 28u + 1$

$Nice quadratic. Let's factorise..$

$27u^2 - 27u - u + 1$

$(27u - 1)(u - 1)$

$Subbing back in u = a^3$

$(27a^3 - 1)(a^3 - 1)$

$oh hey, more difference of two cubes!$

$(3a - 1)(9a^2 + 3a + 1)(a - 1)(a^2 + a + 1)$

$And of course rearranging to make it look presentable$

$(a -1)(3a - 1)(a^2 + a + 1)(9a^2 + 3a + 1)$

EpikHigh · Mar 6, 2012

It's called Latex

_kuroo · Mar 6, 2012

btw, AWESOME and humorous explanation (y)

_kuroo · Mar 6, 2012

actually new problem, i dont understand how $27u^2 - 28u + 1$ becomes $27u^2 - 27u - u + 1$ and how to make it into $(27u - 1)(u - 1)$

nightweaver066;5752207 [tex said:
27u^2 - 28u + 1[/tex]

$Nice quadratic. Let's factorise..$

$27u^2 - 27u - u + 1$

$(27u - 1)(u - 1)$

$Subbing back in u = a^3$

$(27a^3 - 1)(a^3 - 1)$

$oh hey, more difference of two cubes!$

$(3a - 1)(9a^2 + 3a + 1)(a - 1)(a^2 + a + 1)$

$And of course rearranging to make it look presentable$

$(a -1)(3a - 1)(a^2 + a + 1)(9a^2 + 3a + 1)$

D94 · Mar 6, 2012

_kuroo said:
actually new problem, i dont understand how $27u^2 - 28u + 1$ becomes $27u^2 - 27u - u + 1$ and how to make it into $(27u - 1)(u - 1)$

Split up 28u into 27u + u (basic algebra) and then look for a common factor of (27u -1), which is evident in 27u^2 -u = u(27u-1) and -27u - u = -1 (27u-1). Group factors, results in the final answer of (27u - 1)(u - 1).

_kuroo · Mar 6, 2012

ok i figured out why we split, but i dont understand how does grouping factors work?

D94 said:
Split up 28u into 27u + u (basic algebra) and then look for a common factor of (27u -1), which is evident in 27u^2 -u = u(27u-1) and -27u - u = -1 (27u-1). Group factors, results in the final answer of (27u - 1)(u - 1).

nightweaver066 · Mar 6, 2012

_kuroo said:
ok i figured out why we split, but i dont understand how does grouping factors work?

$27u^2 - 28u + 1$

$= 27u^2 - 27u - u + 1$

$= 27u(u - 1) - 1(u - 1)$

$Factorising out (u - 1)$

$= (27u - 1)(u - 1)$

_kuroo · Mar 6, 2012

now that makes sense!

nightweaver066 said:
$27u^2 - 28u + 1$

$= 27u^2 - 27u - u + 1$

$= 27u(u - 1) - 1(u - 1)$

$Factorising out (u - 1)$

$= (27u - 1)(u - 1)$

factorising help (1 Viewer)

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