Question 227394: they want this factored completely
98a^4-72b^2
im not understanding
Found 2 solutions by Alan3354, rapaljer:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Why not try?
98 and 72 are both even numbers, right? Factor out a 2.
= 2(49a^4 - 36b^2)
Both 49 and 36 are perfect squares.
= 2(7a^2 - 6b)(7a^2 + 6b)
Answer by rapaljer(4671)  (Show Source):
You can put this solution on YOUR website! The first step in factoring is ALWAYS to factor the COMMON FACTOR, which in this problem is 2.
98a^4-72b^2
2(49a^4-36b^2)
Notice that you have PERFECT SQUARES inside the parentheses.
2(_______)(_______)
The First times First has to come out to 49a^4. So, try this combination 7a^2*7a^2.
2(7a^2____)(7a^2____)
Now, the Last times Last has to come out to 36b^2. Try 6b*6b.
2(7a^2__6b)(7a^2__6b)
In order to get this to come out to -36b^2, you need OPPOSITE signs:
2(7a^2-6b)(7a^2+6b)
For additional explanation on FACTORING, please see my own website. Do a "Bing" or "Google" search for my last name "Rapalje". Look for "Rapalje Homepage" near the top of these searches. Then near the top of my Homepage, look for "Basic, Intermediate and College Algebra". Choose "Basic Algebra", "Chapter 2". There are several topics on Factoring, especially "Factoring the Common Factor", "Factoring Trinomials", and "Factoring the Difference of Squares." See also the "MATH IN LIVING COLOR" pages that go with these sections.
R^2
Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus
|
|
|
