Math Is Fun Forum

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Cubics and more

Please tell me how to factorise expressions such as

etc...

Also, do they use the same methods, or must I use different methods for each?

I really only know how to solve quadratics and that

has 'n' x intercepts. And that's about all I know xP

thx,
toast

Devantè

Real Member

Registered: 2006-07-14

Posts: 6,400

Re: Cubics and more

http://www.teacherschoice.com.au/Maths_Library/Algebra/Alg_19.htm

If this isn't enough, I'll get back to you with more links.

Devantè

Real Member

Registered: 2006-07-14

Posts: 6,400

Re: Cubics and more

Alternatively, for practice, you can look at some of my exercises in the 'Exercises' forum, which revolve around the topic you are dealing with.

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Re: Cubics and more

Thanks for the help devante, but i don't think that the methods are the same for cubics or quartics .

polylog

Member

Registered: 2006-09-28

Posts: 162

Re: Cubics and more

For the cubics, you need to use the factor theorem, which states:

let P(x) be a cubic

(x - a) is a factor of P(x)   if and only if  P(a) = 0.

so for your example:

You need to try to test some values that will make this equal to 0. Let's try 1, -1, 0, 2, -2 or such.
Usually it is easy since the questions in books are designed to be
I usually try -2 first and look:

-8 + 8 -8 + 8 = 0 success!

Therefore, P(-2) = 0. By the factor theorem:

x -- 2 = (x + 2) is a factor of P(x)

Now you use polynomial long division, divide x + 2 into P(x).

Your result will be a quadratic factor, lets call it Q(x).

Then P(x) = (x + 2)Q(x)

and you can factor Q(x) using the standard methods.

polylog

Member

Registered: 2006-09-28

Posts: 162

Re: Cubics and more

I would assume that quartics (4th degree polynomials) can be factored in the same way, but I haven't done very much of that! You would need to find a factor (x - a), divide it into the quartic, then get a cubic as a result. Then find a factor (x - b) of the cubic by the above process, and divide again to get a quadratic! Then the quartic P(x) = (x - a)(x - b)Q(x) where Q(x) is factored in the usual way.

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Re: Cubics and more

Thanks, I'll try to digest that