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Factoring Trinomials - Difference of Two Squares


Date: 12/22/97 at 22:22:15
From: Matt Church
Subject: Factoring a perfectly square trinomial, and a quadratic 
trinomial

Can you show me how I would  factor the common things in this?  
Especially in the difference of squares, using an example problem 
(because I don't want you to do my homework, because then I wouldn't 
learn anything) and maybe just go throgh it step by step?  If you 
could, it would really help. Thanks.


Date: 12/23/97 at 14:30:55
From: Doctor Rob
Subject: Re: Factoring a perfectly square trinomial, and a quadratic 
trinomial

If you are given an expression you can recognize as the difference
of two squares, you can factor it, because

   u^2 - v^2 = (u - v)*(u + v).

You can verify this by multiplying out the right side using either the
Distributive Law or FOIL. The terms involving u*v cancel each other.

Example: Factor 16*y^6 - z^2/9. Recognize that 16*y^6 = (4*y^3)^2, and 
z^2/9 = (z/3)^2, so this is the difference of two squares. Thus it
must factor, and 16*y^6 - z^2/9 = (4*y^3 - z/3)*(4*y^3 + z/3). (Here 
we were using u = 4*y^3 and v = z/3.)

Next let's learn how to "complete the square."  This means taking a
quadratic trinomial a*x^2 + b*x + c and rewriting it in the form
(A*x + B)^2 + C. If you expand this last form, you will get
A^2*x^2 + 2*A*B*x + (B^2 + C). If this is to equal a*x^2 + b*x + c,
you must have the following three equations:

   A^2 = a,
   2*A*B = b,
   B^2 + C = c.

Solving for A, B, and C, we get:

   A = Sqrt[a],
   B = b/(2*Sqrt[a]),
   C = c - b^2/(4*a) = (4*a*c - b^2)/(4*a).

Example:  Complete the square in 9*x^2 + 6*x - 8:

   a = 9,
   b = 6,
   c = -8,
   A = Sqrt[a] = Sqrt[9] = 3,
   B = b/(2*A) = 6/(2*3) = 1,
   C = c - B^2 = -8 - 1^2 = -9.

Thus 9*x^2 + 6*x - 8 = (3*x + 1)^2 - 9.

If C turns out to be 0, then you are done!  The factorization is 
(A*x+B)^2, and the original polynomial was a perfect square trinomial.
   
Now we are ready to factor quadratic trinomials. Here are the steps:

  1. Complete the square.
  2. Use the difference of squares formula to factor that expression.
  3. Simplify.

To factor 9*x^2 + 6*x - 8, complete the square as above, getting
(3*x + 1)^2 - 9 = (3*x + 1)^2 - 3^2.  This is the difference of 
squares, so it factors as 

  [(3*x+1) - 3]*[(3*x+1) + 3] = (3*x - 2)*(3*x + 4).

That was an easy one, because a = 9 was a square already.  Let's do 
one where it is not a square: 6*z^2 + 13*z - 5. One way to go is to 
multiply and divide by 6 to make the coefficient of z^2 a square:

   6*z^2 + 13*z - 5 = (36*z^2 + 78*z - 30)/6,
                    = [(6*z + 13/2)^2 - 289/4]/6,
                    = [(6*z + 13/2)^2 - (17/2)^2]/6,
                    = [(6*z+13/2) - 17/2)]*[(6*z+13/2) + 17/2)]/6,
                    = (6*z - 2)*(6*z + 15)/6,
                    = (3*z - 1)*(2*z + 5).

Another way to go is to factor out the 6 from all terms and complete 
the square that way:

   6*z^2 + 13*z - 5
      = 6*[z^2 + (13/6)*z - 5/6],
      = 6*[(z + 13/12)^2 - 289/144],
      = 6*[(z + 13/12)^2 - (17/12)^2],
      = 6*(z + 13/12 - 17/12)*(z + 13/12 + 17/12),
      = 6*[(z - 4/12)*(z + 30/12),
      = 6*(z - 1/3)*(z + 5/2),
      = (3*z - 1)*(2*z + 5).

Yet a third way to factor 6*z^2 + 13*z - 5 is to realize that if the
factorization looks like (p*z + q)*(r*z + s), then, multiplying this 
out,

   p*r = 6,
   p*s + q*r = 13,
   q*s = -5.

This means that p must be a positive or negative divisor of 6, and q 
must be a positive or negative divisor of -5. Then r = 6/p and 
s = -5/q.  Make a table of those divisors:

    p   q   r   s   p*s+q*r

    6   5   1  -1     -1
    6   1   1  -5    -29
    6  -1   1   5     29
    6  -5   1   1      1
    3   5   2  -1      7
    3   1   2  -5    -13
    3  -1   2   5     13
    3  -5   2   1     -7
    2   5   3  -1     13
    2   1   3  -5     -7
    2  -1   3   5      7
    2  -5   3   1    -13
    1   5   6  -1     29
    1   1   6  -5      1
    1  -1   6   5     -1
    1  -5   6   1    -29
   -1   5  -6  -1    -29
   -1   1  -6  -5     -1
   -1  -1  -6   5      1
   -1  -5  -6   1     29
   -2   5  -3  -1    -13
   -2   1  -3  -5      7
   -2  -1  -3   5     -7
   -2  -5  -3   1     13
   -3   5  -2  -1     -7
   -3   1  -2  -5     13
   -3  -1  -2   5    -13
   -3  -5  -2   1      7
   -6   5  -1  -1      1
   -6   1  -1  -5     29
   -6  -1  -1   5    -29
   -6  -5  -1   1     -1

Actually, we don't need the bottom half of the table, since we can 
assume that p > 0 (if p < 0, multiply both factors by -1). Throw that 
away. Similarly, we can assume that p >= |r| (if p < |r|, swap the two 
factors), so we don't need the last half of what is left. Throw that 
away, too. That leaves the following table:

    p   q   r   s   p*s+q*r

    6   5   1  -1     -1
    6   1   1  -5    -29
    6  -1   1   5     29
    6  -5   1   1      1
    3   5   2  -1      7
    3   1   2  -5    -13
    3  -1   2   5     13
    3  -5   2   1     -7

Eight rows isn't too bad. Now if 13, the coefficient of z, doesn't 
appear in the last column, then the trinomial won't factor with whole 
number coefficients. If it does appear, it will appear there just 
once.  There is a line with a 13 in the last column. That line tells 
us that p = 3, q = -1, r = 2, s = 5, and the factorization is

   6*z^2 + 13*z - 5 = (3*z - 1)*(2*z + 5).

Some people learn to do this kind of table approach, using the 
divisors of a and c, in their heads. You will not be expected to do 
that, fortunately!

You try some:

   Factor 2*x^2 + 7*x - 4.
   Factor 3*d^2 + 7*d - 6.
   Factor 4*w^4 - 13*w^2 + 3. (Hint: Let x = w^2, so x^2 = w^4.)
   Factor 4*x^2 - 20*e*x + 25*e^2.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Polynomials

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