SOLUTION: find each product university of phoenix elementary and intermediate algebra (3ypower2 + 1) 3y power 2-1)

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Question 93557: find each product university of phoenix elementary and intermediate algebra (3ypower2 + 1) 3y power 2-1)
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
%283y%5E2%2B1%29%283y%5E2-1%29
.
Once you get familiar with this you remember this factoring form:
.
%28a%5E2+-+b%5E2%29+=+%28a+%2B+b%29%28a+-+b%29
.
This form says that if you have the difference of two squares, it factors into the product
of two factors, one the sum of the square roots of the squares and one the difference
of their square roots.
.
What this problem gives you is the sum and difference factors, and you take them back to the
original term by recognizing that you square each term in the factors and subtract them.
.
So if you quickly spot this pattern, you can tell yourself that the square of 3y%5E2 is
9y%5E4 and the square of 1 is 1, so the original expression obtained from
multiplying the two terms you were given is 9y%5E4+-+1.
.
But let's multiply out what you were given to get the product. Start with the given expressions
of:
.
%283y%5E2%2B1%29%283y%5E2-1%29
.
To do this multiplication you take the first factor of %283y%5E2+%2B1%29 and you use (one at a time)
its two terms to multiply the second factor of %283y%5E2+-+1%29. So you begin by taking the
3y%5E2 from the first factor and you use it to multiply the second factor. In other
words you do the following distributed multiplication:
.
3y%5E2%2A%283y%5E2+-+1%29
.
This means you multiply the 3y%5E2 times each of the two terms in parentheses
to get:
.
3y%5E2%2A%283y%5E2+-+1%29+=+9y%5E4+-+3y%5E2
.
and you remember this result.
.
Next you go back to the second term in the first factor ... the +1. And you use that term
to also multiply the second factor. In other words you do the following distributed
multiplication:
.
1%2A%283y%5E2+-+1%29
.
This means you multiply the 1 times each of the two terms in parentheses
to get:
.
1%2A%283y%5E2+-+1%29+=+3y%5E2+-+1
.
Now you combine this result with the previous result you were to remember and you get:
.
9y%5E4+-+3y%5E2+%2B+3y%5E2+-1
.
Notice that the two middle terms are equal but have opposite signs. Therefore, they cancel
each other out and you are left with the answer of:
.
9y%5E4+-1
.
Hope this isn't too confusing to you. Once you see the multiplication pattern it becomes
pretty easy to work problems such as these ... just take the terms in the first factor
and one at a time use them to multiply each of the terms in the second factor. Then
combine the results of all the products that you end up with. This same process will work
for multiplying other types of factors such as multiplying:
.
%28a+%2B+b%29%2A%28c%2Bd%2Be%29 or
+%28a%2Bb%2Bc%29%2A%28d%2Be%2Bf%29 or
+%28a%2Bb%2Bc%29%2A%28d%2Be%2Bf%2Bg%29
.
Just use each of the terms in the first factor to multiply the second factor and when you
are done, just combine all the products you got.
.
Hope also that this helps you to see the logical pattern in multiplying factors.