SOLUTION: please help factor out polynomial. 14x^3y+21x^2y^2-35xy^3 thanksPeacee --Kyle

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Question 191403: please help factor out polynomial.

14x^3y+21x^2y^2-35xy^3

thanksPeacee

--Kyle

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

14x%5E3y%2B21x%5E2y%5E2-35xy%5E3 Start with the given expression


7xy%282x%5E2%2B3xy-5y%5E2%29 Factor out the GCF 7xy


Now let's focus on the inner expression 2x%5E2%2B3xy-5y%5E2




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Looking at 2x%5E2%2B3xy-5y%5E2 we can see that the first term is 2x%5E2 and the last term is -5y%5E2 where the coefficients are 2 and -5 respectively.

Now multiply the first coefficient 2 and the last coefficient -5 to get -10. Now what two numbers multiply to -10 and add to the middle coefficient 3? Let's list all of the factors of -10:



Factors of -10:
1,2,5,10

-1,-2,-5,-10 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to -10
(1)*(-10)
(2)*(-5)
(-1)*(10)
(-2)*(5)

note: remember, the product of a negative and a positive number is a negative number


Now which of these pairs add to 3? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 3

First NumberSecond NumberSum
1-101+(-10)=-9
2-52+(-5)=-3
-110-1+10=9
-25-2+5=3



From this list we can see that -2 and 5 add up to 3 and multiply to -10


Now looking at the expression 2x%5E2%2B3xy-5y%5E2, replace 3xy with -2xy%2B5xy (notice -2xy%2B5xy adds up to 3xy. So it is equivalent to 3xy)

2x%5E2%2Bhighlight%28-2xy%2B5xy%29%2B-5y%5E2


Now let's factor 2x%5E2-2xy%2B5xy-5y%5E2 by grouping:


%282x%5E2-2xy%29%2B%285xy-5y%5E2%29 Group like terms


2x%28x-y%29%2B5y%28x-y%29 Factor out the GCF of 2x out of the first group. Factor out the GCF of 5y out of the second group


%282x%2B5y%29%28x-y%29 Since we have a common term of x-y, we can combine like terms

So 2x%5E2-2xy%2B5xy-5y%5E2 factors to %282x%2B5y%29%28x-y%29


So this also means that 2x%5E2%2B3xy-5y%5E2 factors to %282x%2B5y%29%28x-y%29 (since 2x%5E2%2B3xy-5y%5E2 is equivalent to 2x%5E2-2xy%2B5xy-5y%5E2)



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So our expression goes from 7xy%282x%5E2%2B3xy-5y%5E2%29 and factors further to 7xy%282x%2B5y%29%28x-y%29


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Answer:

So 14x%5E3y%2B21x%5E2y%5E2-35xy%5E3 completely factors to 7xy%282x%2B5y%29%28x-y%29