Can some one please factor (x^3 - 8)

[whatdofisheat]

ya title is pretty much all i need (x^3 - 8)
if you could factor that it would be of great help

 

[dextercioby]

HINT:8=2^{3}


Daniel.

 

[Jameson]

I assume you know the formulas for the sum / difference of cubes, since your homework is asking a question that pertains to this method of factoring.

Rewrite your problem as (x^3-2^3)

Can you see it now?

Jameson

EDIT: In case you don't have the formula, I'll be nice... here you go.

(x^3-y^3) = (x-y)(x^2+xy+y^2)

 

[whatdofisheat]

thanks for the formula i have never seen that before
but another methode we are trying to use is synthetic division
if anyone can do it that way it would also help
thanks
fish

 

[dextercioby]

x^{3} x^{2} x^{1} x^{0}
coeff. 1 0 0 -8
2 1 2 4 0​

Solution

x^{3}-8=(x-2)\left(x^{2}+2x+4\right)

Daniel.

 

[uart]

thanks for the formula i have never seen that before
but another methode we are trying to use is synthetic division
if anyone can do it that way it would also help
thanks
fish

Yes well if you don't know the formula for factorization of difference of two cubes then polynomial division is a good way to proceed.

In order to use the division method you must first obtain one factor by some means, possibly guess. With the difference of two cubes, x^3 - a^3, it's very easy to see that x=a is a zero and hence (x-a) is a factor. So essentially you obtain this first factor by inspection in this case.

Now just do the polynoimial division (x^3 - a^3) / (x-a) to obtain the other less obvious factor.

 

[whatdofisheat]

thanks for all your help
i got it now

 

[mathwonk]

A basic result once taught early in high school, and called the "root-factor theorem", is that whenever x=a makes a polynomial equal to zero (i.e. if a is a "root"), then x-a is a factor of that polynomial.

For some reason this fact seems to be unknown to most first year college calculus students today.

 

[dextercioby]

I wonder which is more important to know when u graduate HS:the root factor theorem (why this theorem and not others) or

a^{3}-b^{3}=(a-b)\left(a^{2}+ab+b^{2}\right)

Daniel.