# Doodles Do Algebra – Lesson 51

Update: This Lesson Is Part of Book 3 of Doodles Do Algebra: “The Basic Math of Algebra” available on Amazon.

Worksheet:

Available in the book, “The Basic Math of Algebra, Book 3 of Doodles Do Algebra“

Today your child learns to multiply two polynomials. It is set up much the same as she learned to multiply 2 digit numbers, only you don’t have to keep track of place value.

The basic idea is that each term of the first polynomial gets a chance to multiply each term in the second polynomial. Stacking the polynomials is something I never saw growing up, but my husband who is a bit older did do polynomial multiplication this way, so it is a technique used from the 1700’s through at least the early 1960’s. My kids really understood much better this way than the “memorize the property and apply it” method. This style of learning algebra also fits if you are teaching with classical methods since at the stage which your child would be learning algebra, they are most likely at a “logic” stage of development at which point rote memorization is actually retarding their intellectual growth (I know this sounds kind of snooty, but it is really not. I strongly feel that rote memorization is not learning once your child is older than about 10.)

So, the first problem your child works today is also shown as an example work through by DoodleTwo. The process is to multiply the first term (we use d in this example) by the top polynomial (or a+b), one term at a time. This yields ad + bd (remember your hippo hopping here!). Then you do the same process with the second term, c in this example which yields ac + cb, which you add to the first resultant. And so the whole answer is the addition of all these terms, or ad + bd + ac + cb.

Problem 1. $ad + bd + ac + cb$
Problem 2. Step one is to multiply 2x + 3y by 2b, and that gets you (2)(b)(2)(x) + (3)(y)(2)(b) = 4bx + 6by. Step two is to add the contribution of multiplying 2x+3y by the second term, in this case 3a {note that you can do this in any order you want, you just need to make sure you multiply everything through and add it all together}, and that gets you (2)(x)(3)(a) + (3)(y)(3)(a) = 6ax+9ay. The final step is to add all the terms together which gets you $4bx+6by + 6ax+9ay$. In this case the terms are all different enough that they cannot be added together to reduce your answer. So that is it.