hey can someone pls help me again

Question

here is the the link that has the pic of what I need help with:

help please, sorry, i know its kind of alot - Brainly.com

Answer 1

First off let me be honest. This teaching method is VERY different than how I was told twenty years ago. I think I have it figured out though.

For the "product" portions add the top right and bottom left pieces. For the factors you can either work from the product or from the cube. I honestly prefer working from the cube.

The way I have figured out these problems is somewhat simple. Take either the top or bottom row and figure out what the common denominator is. For example in 5 the common denominator of the bottom is 5 (funny how that works out.). That would make the bottom 5(x+2). If you multiply through you'll get the two pieces of the bottom row. Now you need to divide the top by one of the pieces from the bottom {5(X+2)}. Seeing as all the numbers in the top row are less than five it can't be the five, so it must be the x+2. I you divide 2x^2+4x by X+2 you get 2x. You then combine that with the leftover from the bottom row (5). That gives you (2x+5)(x+2). If you do FOIL (not sure if that's still taught, but if not it's First, Outer, Inner, Last) that gives 2x^2+4x+5x+10 or 2x^2+9x +10. The same answer you should have gotten from adding the top right and bottom left squares.

Lets try the next one. First we add up the top right and bottom left to get the product or X^2 + 12x+32. I you can solve from there... Great! If not let's continue with our original plan. You can chose either top or bottom, but the top seems simplest so lets go with it. we can factor it to x(x+4). Now we look at the bottom row. There is no x in the bottom right, so it must be able to divide by (x+4).

We can! The answer is 8(x+4). Just like last time we join the two pieces left over to make (x+8)(x+4). If you double check the answer then you'll find that equation also yields x^2 + 12x + 32.

Hopefully this has helped. If not I'll work through a few more with you.

Best,

Dr.D