Certifictae in further mathematics (factorising quadratics with cubed unknowns)

Okay so I am studying the aqa certificate in further mathematics and I would like some help. I don't understand how to factorise quadratics which also have cubed unknowns (e.g. x^3 +x^2 +2x +1). I know that they are meant to have three brackets but I have no idea how to figure out what should be in the brackets. My class has a test on Wednesday January 6th and I am worried that this might come up on the test. I've been looking online and I haven't found anything that explains how to do it so I would very appreciative if someone could help me.
Thanks for reading and more thanks if you can help.
:smile:

Reply 1

Duke Glacia

16

Original post by AspiringUnderdog

Okay so I am studying the aqa certificate in further mathematics and I would like some help. I don't understand how to factorise quadratics which also have cubed unknowns (e.g. x^3 +x^2 +2x +1). I know that they are meant to have three brackets but I have no idea how to figure out what should be in the brackets. My class has a test on Wednesday January 6th and I am worried that this might come up on the test. I've been looking online and I haven't found anything that explains how to do it so I would very appreciative if someone could help me.
Thanks for reading and more thanks if you can help.
:smile:

theres no exact way to factorise a cubic equation(unlike quadratic eqn ). so try numbers like x=1,2,3,5.-1,-2,-3,-5,-7 and use factor theorum to factorise

Reply 2

Kevin De Bruyne

21

Original post by AspiringUnderdog

Okay so I am studying the aqa certificate in further mathematics and I would like some help. I don't understand how to factorise quadratics which also have cubed unknowns (e.g. x^3 +x^2 +2x +1). I know that they are meant to have three brackets but I have no idea how to figure out what should be in the brackets. My class has a test on Wednesday January 6th and I am worried that this might come up on the test. I've been looking online and I haven't found anything that explains how to do it so I would very appreciative if someone could help me.
Thanks for reading and more thanks if you can help.
:smile:

Have you come across the factor theorem?

The example you've provided isn't factorisable, so if we take something that is, eg x^3 + x^2 - x - 1.

The factor theorem tells us that if, some number a, is a root (i.e f(a) = 0) then it can be written as f(x) = (x-a)(q(x)) where q(x) is some other function, in this case a quadratic, that you have to find and solve.

Eg if 5 was a root of that equation, then x^3 + x^2 - x - 1 = (x-5)(q(x)). You then have to use other things that you know to find q(x). Hint: (x^3 + x^2 - x - 1)/(x-5) = q(x). But 5 isn't necessarily a root.

And how do you find a root? Well, see if you can spot it from the equation.

Reply 3

Johann von Gauss

14

Original post by Duke Glacia

theres no exact way to factorise a cubic equation(unlike quadratic eqn ). so try numbers like x=1,2,3,5.-1,-2,-3,-5,-7 and use factor theorum to factorise

erm... not strictly true...

a x^3 + b x^2 + c x + d = 0 \implies

But none (in radicals) for 5 order and above IIRC

https://en.wikipedia.org/wiki/Galois_theory#A_non-solvable_quintic_example

Edit: OP, do NOT try to remember this formula! :tongue:

(edited 8 years ago)

Reply 4

Duke Glacia

16

Original post by Johann von Gauss

erm... not strictly true...

a x^3 + b x^2 + c x + d = 0 \implies

But none for 5 order and above IIRC

wtf is that.................................... :eek4:

Reply 5

Gregorius

14

Original post by Duke Glacia

wtf is that.................................... :eek4:

You might find the Wikipedia articles on the cubic and on the quartic a bit more illuminating, as they unwrap the process of solving such equations by radicals. Beyond the quartic, whether a given polynomial equation is solvable by radicals depends on whether a particular group (that basically arises as a permutation group of the roots of the equation) is solvable.

For the quintic and beyond, one should note that they are solvable in terms of special functions called modular functions. Alas, this fascinating theory is rarely taught even at undergraduate level. :colonhash:

Reply 6

Johann von Gauss

14

Original post by Gregorius

You might find the Wikipedia articles on the cubic and on the quartic a bit more illuminating, as they unwrap the process of solving such equations by radicals. Beyond the quartic, whether a given polynomial equation is solvable by radicals depends on whether a particular group (that basically arises as a permutation group of the roots of the equation) is solvable.

For the quintic and beyond, one should note that they are solvable in terms of special functions called modular functions. Alas, this fascinating theory is rarely taught even at undergraduate level. :colonhash:

Can't wait to learn Galois Theory :smile:

Aren't modular functions quite new? IIRC, they were involved in Wiles's proof of Fermat's Last Theorm

Reply 7

Gregorius

14

Original post by Johann von Gauss

Can't wait to learn Galois Theory :smile:

Aren't modular functions quite new? IIRC, they were involved in Wiles's proof of Fermat's Last Theorm

Galois theory is just lovely; I hope you enjoy it when you arrive :smile:

Modular forms and functions are most definitely from the golden age of nineteenth century mathematics; if you pop down to the history section of the Wikipedia article, you'll see that Wiles's work lies very much in the "great tradition" that grew out of Euler, Gauss, Jacobi...

Reply 8

AspiringUnderdog

OP

21

Okay I found a way to do it, thanks for the help, guys. :smile:

Reply 9

B_9710

18

Original post by AspiringUnderdog

Okay I found a way to do it, thanks for the help, guys. :smile:

It may be worth remembering something called the rational roots theorem, which basically states that if any rational roots of a polynomial exist, in this case a cubic(this also implies that the cubic can be factorised into linear factors - if solutions exist) then they are are one of the factors of the constant term divided by one of the factors of the leading coefficient - the coefficient of x^3 term in a cubic equation. So if you have for example 2x^3+3x^2+4x+4=0. The rational roots (if any exist) are of the form +/- (1,2,4)/(1,2) ie. +/- 1, +/- 1/2, +/- 2, +/- 4.
The reason I'm pointing this out is because I think it's important to bear in mind when using the factor theorem so you know which numbers to try. For GCSE though they will give you 'nice cubics, with 3 rational (real roots). I probably haven't explained this too well but it may be worth looking at the Wikipedia page on the rational roots theorem.

Certifictae in further mathematics (factorising quadratics with cubed unknowns)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you