Polynomial Factoring Rules

Date: 04/02/97 at 00:22:03
From: Natasha Price
Subject: Polynomial Factoring Rules

Hello.  I have a problem with the "special cases" of factoring 
polynomials - can you help? 

1)  t^21 + 1
2)  25y^2-144 = 0

Date: 04/02/97 at 12:17:50
From: Doctor Wilkinson
Subject: Re: Polynomial Factoring Rules

You probably know the rules

 a^2 - b^2 = (a + b) (a - b)

 a^3 - b^3 = (a - b) (a^2 + ab + b^2)

 a^3 + b^3 = (a + b) (a^2 - ab + b^2)

It looks like your problem is in recognizing when these rules apply. 

Let's take your first question:

 t^21 + 1.

The only rule which looks like it might work is the third one above.  
If we're going to apply it, we need to be able to express t^21 as the 
cube of something (a in the rule as I've given it), and 1 as the cube 
of something (b in the rule).  Can we do this?  Well fortunately the 
exponent 21 is a multiple of 3, namely 3 * 7, so by the rules of 
exponents 

 (t^7)^3 = t^(7 * 3) = t^21

so we can write t^21 = (t^7)^3 and t^7 can fit in the place of a in 
the factoring rule.

 1 is no problem, since 1 = 1^3

So we have

 t^21 + 1 = (t^7)^3 + 1^3 = (t^7 + 1) ((t^7)^2 - t^7 + 1)

applying the rule with a = t^7 and b = 1.

This can be simplified to

 (t^7 + 1) (t^14 - t^7 + 1)


Now let's take your next problem, which asks you to solve

 25y^2 - 144 = 0

As you've probably guessed, you're going to solve this by factoring.

This time the rule that we can hope to apply is the first rule, so we 
have to write

 25y^2 - 144

as a difference of two squares.

Fortunately, we have 25 = 5^2, so 25y^2 = (5y)^2; and 144 = 12^2.  
So we can write

 25y^2 - 144 = (5y)^2 = 12^2

and we can apply the first rule with a = 25y and b = 12 to get

 25y^2 - 144 = (5y + 12) (5y -12)

and we set this equal to zero:

 (5y + 12) (5y - 12) = 0

Now the only way to get a product to come out zero is for one of the 
factors to be zero, so to satisfy the equation we have to have either

 5y + 12 = 0

or

 5y - 12 = 0

so that y = -12/5 and y = 12/5 are the solutions.

-Doctor Wilkinson,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/