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 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/
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