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Factoring 8 - t^3


Date: 01/07/98 at 19:53:11
From: Adam Jones
Subject: Sum and difference of cubes

My question is how to do this problem: 

Factor the given expression completely: 8 - t^3. 


Date: 01/12/98 at 11:51:56
From: Doctor Joe
Subject: Re: Sum and difference of cubes

Dear Adam,

Let's start by looking at the following algebraic identity:

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

To see that this is true, we start by expanding the expression on the 
right:

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

Seeing that 8 = 2^3, we write x = 2 and y = t.  Then, we have:

             8 - t^3
           = (2)^3 - (t)^3
           = (2 - t)(2^2 + 2t + t^2)
           = (2 - t)(4 + 2t + t^2)

But at this stage, you must be wondering: "How did you know that 
(x^3 - y^3) has a factor (x - y)?"

Let's explore the real reason behind this clever guess.

The guess lies in the following theorem (without it you would think I 
was very lucky).

Theorem (Remainder/Factor Theorem)

Let f(t) and d(t) be two polynomials such that the 
deg(f(t)) < deg(d(t)).

Polynomials are things that look like this:

  f(t) = 2t^3 + 4t
  g(t) = -2t^7 + 4t^6 - 9t^5 + 40)

and deg(polynomial) (read as the degree of polynomial) denotes the 
highest power of the indeterminate occurring in the polynomial, e.g.:

  deg (2t) = 1, deg (-4t^2 + 5t + 5) = 2)

Then there exists a polynomial q(x) and r(x) such that 
deg(r(x))< deg(d(x)) or r(x) is identically equal to zero and

            f(t) = q(t)d(t) + r(t).

*** End of Theorem ***


Now, to apply this theorem (without proof) to this problem, we
let f(t) = 8 - t^3.

Observe that f(2) = 8 - 2^3 = 8 - 8 = 0.

Let's divide f(t) by (t - 2).  According to the Theorem above, there 
exist a q(t) and r(t) such that deg(r(t))< deg(t-2) = 1 (meaning that 
r(t) is actually a constant number, denoted by r) with

         f(t) = 8 - t^3 = (t- 2)q(t) + r

The above statement is really an identity; i.e. the equality holds for 
all values of t.

In this case we choose t = 2 - now you know why (t -2) is chosen -

         f(2) = 0 = (2 - 2)q(2) + r

implying that r = f(2) = 0.

So, (8 - t^3) leaves no remainder when it is divided by (t -2). Take 
note of the argument: there is no stage where long division is 
actually carried out to test whether (t -2) is a factor of f(t)!

Now we know that 8 - t^3 = (t - 2)q(t) for some q(t). We apply the 
long division algorithm to find out the explicit form for q(t):

                             -t^2 - 2*t - 4
                    ------------------------
             t - 2 | -t^3 + 0*t^2 + 0*t + 8
                     -t^3 + 2*t^2
                     ------------
                           -2*t^2 + 0*t
                           -2*t^2 + 4*t
                           ------------
                                   -4*t + 8
                                   -4*t + 8
                                   --------
                                          0
                                   --------

Hence, 8 - t^3 = (t - 2)(-t^2 - 2t - 4)
               = (2 - t)(t^2 + 2t + 4)

as desired.

Now, test yourself by factorising completely 27 + t^3.

All the best.

-Doctor Joe,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Polynomials
Middle School Algebra
Middle School Factoring Expressions

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