Factoring Quadratic TrinomialsDate: 06/02/98 at 01:42:30 From: Anonymous Subject: Factoring I need to learn how to factor correctly. I sort of have it, but not all the way. Can you help me please? I'm trying to figure out how to factor y^2 + 8y + 16. How do you know how to correctly find the greatest common factor and factor it? Date: 06/02/98 at 09:13:05 From: Doctor Rob Subject: Re: Factoring In general, you will be trying to factor quadratic trinomials that look like this: a*y^2 + b*y + c where a, b, and c are expressions not involving y. In your case above, a, b, and c are constants, 1, 8, and 16, respectively. If this polynomial factors, it will have two factors that look like this: (r*y + s)(t*y + u) Multiplying this out, and setting the coefficients of y^2, y^1 = y, and y^0 = 1, from the two polynomials equal, you get three equations: r*t = a r*u + s*t = b s*u = c You want to find r, s, t, and u to make these equations true. Now: a*c = (r*t)*(s*u) = (r*u)*(s*t) b = (r*u) + (s*t) This means that you are looking for two numbers d = r*u and s*t such that their product is a*c and their sum is b. In your case, you are looking for two numbers whose product is 1*16 = 16, and whose sum is 8. Now run through all positive and negative factors d of a*c, and write down d and (a*c)/d. Pick out the one (if any) such that d and (a*c)/d add up to b. In your case, here is the table: d 16/d sum -16 -1 -17 -8 -2 -10 -4 -4 -8 -2 -8 -10 -1 -16 -17 1 16 17 2 8 10 4 4 8 8 2 10 16 1 17 There are some shortcuts available. First of all, you might as well choose |d| <= |a*c/d|, so that d is the smaller in absolute value of the two numbers for which you are looking. Secondly, d must have the same sign as a*b*c. That will cut down the number of rows in the table by a factor of nearly 4. In your case, only the 6th, 7th, and 8th rows survive. The one which works is 16 = 4*4, 4 + 4 = 8. If there are none which work, then (if you have done this correctly) the polynomial cannot be factored. Now write: a*y^2 + b*y + c = r*t*y^2 + (r*u + s*t)*y + s*u (substituting a, b, c) = r*t*y^2 + r*u*y + s*t*y + s*u (expanding) = (r*t*y^2 + r*u*y) + (s*t*y + s*u) (grouping) = r*y*(t*y + u) + s*(t*y + u) (removing common factors) = (r*y + s)*(t*y + u) (removing a common factor) In your case, y^2 + 8*y + 16 = y^2 + (4 + 4)*y + 16 = y^2 + 4*y + 4*y + 16 = (y^2 + 4*y) + (4*y + 16) = y*(y + 4) + 4*(y + 4) = (y + 4)*(y + 4) = (y + 4)^2 You try to do these: 1. Factor y^2 + 10*y + 25. 2. Factor y^2 - 7*y + 12. 3. Factor y^2 - 7*y - 18. 4. Factor 2*y^2 + 7*y - 4. 5. Factor y^2 + y + 1. -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- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/