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Question 76364: I would like help completely factoring 28ab^2c-56a Thanks!
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! 28ab^2c-56a
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The first thing to do is to look at the constants in each of the two terms to see if they
have any common factors that can be pulled out. In this problem 2 would be a common
factor of both 28 and 56. You can go further and say that 7 is a common factor of both
28 and 56. And if you continue along this line you will see that 14 is a common factor,
and finally 28 is a common factor of both 28 and 56. If you pull the 28 out as a common
factor the result is:
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28*(ab^2c - 2a)
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As a check, notice that if you perform this distributed multiplication you get back to
the original polynomial of the problem.
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Now look at the terms in the parentheses. Are there are any common factors that appear
in both terms. Yes ... the common variable is "a". Pull the "a" out of both the terms
and you get:
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28*a*(b^2c - 2)
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There are no more common variables inside the parentheses, so this is as far as you can
go.
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So the answer to this problem is that the factored form is 28a(b^2c - 2). Again, if you
perform this distributed multiplication you get back to the original polynomial.
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Hope this helps you to see a basic method of factoring polynomials.
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