Factoring problem - is it an iteration or more obvious?

[Sparky_]

Homework Statement

A part of a problem is to factor 8 - b^3

"8 minus b cubed"

Homework Equations

The Attempt at a Solution

I see that the problem is 2^3 - b^3.

I don't see the next step

I know the answer is (2-b)(4+2b+b^2).

but I can't get there.

Is this just trail and error dividing terms into the polynomial or is there a more obvious solution?

Thanks
Sparky_

 

[dynamicsolo]

Are you familiar with polynomial division? (I'm assuming you want an answer better than 'use the difference of two cubes formula'.)

You would look for a value of b which solves the equation
8 - (b^3) = 0 , which gives b^3 = 8 and thus b = 2. The factors of a polynomials can be expressed as differences of the variable and the zeroes of the polynomials (are you familiar with this theorem?). So we now know that one factor of 8 - (b^3) is
(b - 2) . If you divide this factor into the original polynomial, you will get - ( 4 + 2b + b^2 ).

 

[HallsofIvy]

Homework Statement

A part of a problem is to factor 8 - b^3

"8 minus b cubed"

Homework Equations

The Attempt at a Solution

I see that the problem is 2^3 - b^3.

I don't see the next step

I know the answer is (2-b)(4+2b+b^2).

but I can't get there.

Is this just trail and error dividing terms into the polynomial or is there a more obvious solution?

Thanks
Sparky_

(Intelligent) trial and error is pretty much the way you (try to) factor most polynomials (I say "try to" because, of course, most polynomials can't be factored using only integer coefficients).

The other way is to memorize some basic formulas. In particular, a^n- b^n= (a- b)(a^{n-1}+ a^{n-2}b+ a^{n-3}b^2+ \cdot\cdot\cdot+ a^2b^{n-3}+ ab^{n-2}+ b^{n-1}) is useful for this problem. It is also true that if n is odd then a^n+ b^n= (a+b)(a^{n-1}- ba^{n-2}+ b^2a^{n-3}- \cdot\cdot\cdot+a^2b^{n-3}- ab^{n-2}+ b^{n-1}).

Of course, if you don't require that the coefficients be integer, there is a sure-fire method of factoring any polynomial, p(x). First find all roots of the equation p(x)= 0, say x_1, x_2, \cdot\cdot\cdot , x_n where n is the degree of the polynomial, including complex roots and counting the correct multiplicity for each root. Then, if the leading coefficient is a, p(x)= (x- x_1)(x- x_2)\cdot\cdot\cdot(x- x_{n-1})(x-x_n). Of course, that method is not terribly useful if you want to factor in order to solve the equation!

 

[tiny-tim]

… is there a more obvious solution?

Hi Sparky!

Substitute b = 2a.

Then you have 8(1 - a³).

Do you know how to factor 1 - a³ ?