Factoring Trinomials - Difference of Two SquaresDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/