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Complete the Square and Factor


Date: 01/14/97 at 15:11:18
From: darrellnanetteryankody
Subject: Algebra

Howdy from halfway,

I am a 39 year old senior taking my last 3 hours before I graduate in 
a three-week course. I have not taken any previous math. The course is 
College Algebra and since there is an intermediate course before this 
which I did not take, I am lost. I understand the concepts but I am 
not clear on working out the problems. If you could tell me how to 
complete the square and factor, I think it could be very beneficial to 
me. 

Thanks, 
Nanette Holley


Date: 01/14/97 at 17:23:57
From: Doctor Rob
Subject: Re: Algebra

These two topics are closely related.  Let's start with completing the
square.

You know that:

(x+1)^2 = x^2 + 2*x + 1
(x+2)^2 = x^2 + 4*x + 4
(x+3)^2 = x^2 + 6*x + 9
(x+4)^2 = x^2 + 8*x + 16
(x+5)^2 = x^2 + 10*x + 25

In general:

(x+n)^2 = x^2 + 2*n*x + n^2

This works regardless of whether n is even or odd.

When you are faced with a quadratic polynomial in x, you can always
write it as a constant times a square plus another constant, as
follows.  Suppose a, b, and c are some constants which you are given.
(They may actually be expressions in some other variables, but they
must be independent of x!)  You are then given the polynomial below
and asked to "complete the square":

a*x^2 + b*x + c

First, factor out the "a" from *all three* terms:

a*[x^2 + (b/a)*x + (c/a)]

Now use the general fact above to see what square starts off
x^2 + (b/a)*x + ...  You can see that you will need to have 2*n = b/a, 
or n = b/(2*a).  Then (x + b/(2*a))^2 = x^2 + (b/a)*x + b^2/(4*a^2).  
This is the square you want to use, so stick it inside the [.] in the 
above expression.  You do this by adding and subtracting b^2/(4*a^2):

a*[x^2 + (b/a)*x + b^2/(4*a^2) - b^2/(4*a^2) + (c/a)]
= a*[(x + b/(2*a))^2 - b^2/(4*a^2) + (c/a)]

Now all the x-terms are included in the square part.  Combine the
rest of the stuff inside [.] to get the form:

a*(x + b/(2*a))^2 - (b^2 - 4*a*c)/(4*a)

You have completed the square!

Example:  7*x^2 - 11*x + 24
  = 7*[x^2 - (11/7)*x + (24/7)]
  = 7*[(x - 11/14)^2 + 24/7 - (11/14)^2]
  = 7*[(x - 11/14)^2 + (672 - 121)/14^2]
  = 7*[(x - 11/14)^2 + 551/196]

If the coefficient of x^2, which we called a, is itself a square, a 
further simplification is possible.  Say a = r^2.  Then:

a*(x + b/(2*a))^2 - (b^2 - 4*a*c)/(4*a)
 = r^2*[(x + b/(2*a))^2 - (b^2 - 4*a*c)/(4*a^2)]
 = r^2*(x + b/(2*a))^2 - r^2*(b^2 - 4*a*c)/(4*a^2)
 = (r*x + b/(2*r))^2 - (b^2 - 4*a*c)/(4*a)

Example:  36*x^2 - 96*x + 55
 = 6^2*[(x - 96/72)^2 + 55/36 - (96/72)^2]
 = 6^2*[(x - 4/3)^2 + 55/36 - (4/3)^2]
 = (6*x - 8)^2 + 55 - 64
 = (6*x - 8)^2 - 9

--------------------------------------------------------------------

Now as to factoring...

The usual case here is that you are given some quadratic polynomial
in some variable which we will call x.  (It is also possible to have a 
quadratic polynomial in x^2, or 5*x^2+4, or any other expression, but 
we will restrict our attention to x.)  It will have, as above,
coefficients which we will call a, b, and c, which are either 
constants, or expressions which do not involve x.  You will be asked
to factor it into two linear polynomials in x:

a*x^2 + b*x + c = (r*x + s)*(t*x + u)

Again, r, s, t, and u are constants, or expressions not using the 
variable x.  If you multiply out the right side, you will get
(r*t)*x^2 + (r*u + s*t)*x + (s*u).  Setting coefficients of x^2 equal, 
coefficients of x equal, and constant terms equal, we see that:

a = r*t
b = r*u + s*t
c = s*u

The trick is to find r, s, t, and u to make these equations true.  The 
first observation is that if a, b, and c are integers, r, s, t, and u 
will also be integers.  The second observation is that r and t must be 
divisors of a, and s and u must be divisors of c.  Now you can go down 
the list of divisors of a for r and the divisors of c for s, compute t 
and u, and check b = r*u + s*t.  This will give you an answer if there 
is one.

Another approach is to use completing the square (I told you that
these were related!).  Complete the square for this polynomial:

a*x^2 + b*x + c = a*[(x + b/(2*a))^2 - (b^2 - 4*a*c)/(4*a^2)]

Now use the fact that you can factor A^2 - B^2 = (A + B)*(A - B) to 
factor the expression inside [.].  This will be possible if you can 
write down a d such that d^2 = b^2 - 4*a*c.  If this expression is a 
perfect square, you are in business.  If not, you can't factor the 
original polynomial without using square roots like Sqrt[2] or 
Sqrt[-17].  The inside part now looks like:

(x + b/(2*a))^2 - (d/(2*a))^2 = A^2 - B^2

where A = x + b/(2*a) and B = d/(2*a).  The factorization now looks
like:
  
a*[x + b/(2*a) + d/(2*a)]*[x + b/(2*a) - d/(2*a)]
 = a*[x + (b + d)/(2*a)]*[x + (b - d)/(2*a)]

Once you reduce (b + d)/(2*a) and (b - d)/(2*a) to lowest terms, you 
will find the denominators are your sought-after numbers r and t, 
respectively, and the numerators are s and u, respectively.

Example:  Factor 36*x^2 - 96*x + 55

36*x^2 - 96*x + 55
 = (6*x - 8)^2 - 9, as we found above, and 9 = 3^2, so d = 3
 = (6*x - 8 + 3)*(6*x - 8 - 3)
 = (6*x - 5)*(6*x - 11)

Here r = 6, s = -5, t = 6, and u = -11, and the three equations are 
satisfied.

Example:  Factor 5*x^2 + 32*x - 21
 = 5*[(x + 16/5)^2 - 21/5 - 256/25]
 = 5*[(x + 16/5)^2 - 361/25]

Here 361/25 = (19/5)^2, so d = 19/5

   5*x^2 + 32*x - 21
 = 5*[x + 16/5 + 19/5]*[x + 16/5 - 19/5]
 = 5*[x + 35/5]*[x - 3/5]
 = 5*[x + 7/1]*[x - 3/5]
 = (x + 7)*(5*x - 3)

We could also have tried r = -5, -1, 1, and 5, and s = -21, -7, -3, 
-1, 1, 3, 7, and 21, to see which ones satsified 32 = -21*r/s + 5*s/r,
finally finding that r = 1 and s = 7 work or that r = 5 and s = -3
work, to achieve the same result.

I hope this helps.  If not, write back to us again.

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

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