Q&A #6091

Teachers' Lounge Discussion: Monomials and polynomials

T2T || FAQ || Ask T2T || Teachers' Lounge || Browse || Search || T2T Associates || About T2T

View entire discussion
[<< prev] [ next >>]

From: Loyd

To: Teacher2Teacher Public Discussion
Date: 2005102611:26:20
Subject: Re: Alegbra 1AB

On 2005102520:59:57, nick wrote: > >Im in 7th grade and learning about monomials and polynomials. My >teacher is great and all, and i do understand, but the way he taught >it to our class take forever to do the simplest of problems. I wanted >to know the easiest way to simplify an equation where you multiply a >monomial by a monomial, a monomial by a polynomial, and especially a >polynomial by a polynomial. Thanks a bunch!!! > > > :)Nick:) > How about (a+b)(c+d) You can multiply a times every term in the second polynomial then do the same thing with b: ac+ad+bc+bd. That is the way I usually do it: But you can use the distributive law: Multiply the first binomial times every term in the second binomial: (a+b)c + (a+b)d= ac + bc + ad+bd. Either way, you get the same answer after rearranging terms. You can use either method to do the same for trinomials. Just a few more terms to confuse you, possibly. You can make up polynomial you know the answer to if you have trouble: (1+2+3+4)(2 + 3 +7) = 10x12. One can see the answer is 120. Using * for multiply: 1*2 + 1*3+1*7 + 2*2+2*3+2*7 +3*2+3*3+3*7+4*2+4*3+4*7 2 + 3 + 7 + 4 + 6 +14 +6+9 +21 +8 +12 +28 =120 I only made to mistakes that I had to correct. As you can see, I didn't have much to do today.

Post a reply to this message
Post a related public discussion message
Ask Teacher2Teacher a new question

[Privacy Policy] [Terms of Use]

Math Forum Home || The Math Library || Quick Reference || Math Forum Search

Teacher2Teacher - T2T ®
© 1994- The Math Forum at NCTM. All rights reserved.