SOLUTION: Factor the polynomial completly: x^3y+2x^2y^2+xy^3 Factor the polinomial completly: 32x^2y-2y^3 Use grouping to factor the polynomial completly: x^3+x^2-x-1 Use g

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor the polynomial completly: x^3y+2x^2y^2+xy^3 Factor the polinomial completly: 32x^2y-2y^3 Use grouping to factor the polynomial completly: x^3+x^2-x-1 Use g      Log On


   

Question 188463: Factor the polynomial completly:
x^3y+2x^2y^2+xy^3
Factor the polinomial completly:
32x^2y-2y^3
Use grouping to factor the polynomial completly:
x^3+x^2-x-1
Use grouping to factor the polynomial completly"
y^3-5y^2+8y-40
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first two to get you started

# 1




x%5E3y%2B2x%5E2y%5E2%2Bxy%5E3 Start with the given expression


xy%28x%5E2%2B2xy%2By%5E2%29 Factor out the GCF xy


Now let's focus on the inner expression x%5E2%2B2xy%2By%5E2




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Looking at x%5E2%2B2xy%2By%5E2 we can see that the first term is x%5E2 and the last term is y%5E2 where the coefficients are 1 and 1 respectively.

Now multiply the first coefficient 1 and the last coefficient 1 to get 1. Now what two numbers multiply to 1 and add to the middle coefficient 2? Let's list all of the factors of 1:



Factors of 1:
1

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

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

note: remember two negative numbers multiplied together make a positive number


Now which of these pairs add to 2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 2

First NumberSecond NumberSum
111+1=2
-1-1-1+(-1)=-2



From this list we can see that 1 and 1 add up to 2 and multiply to 1


Now looking at the expression x%5E2%2B2xy%2By%5E2, replace 2xy with xy%2Bxy (notice xy%2Bxy adds up to 2xy. So it is equivalent to 2xy)

x%5E2%2Bhighlight%28xy%2Bxy%29%2By%5E2


Now let's factor x%5E2%2Bxy%2Bxy%2By%5E2 by grouping:


%28x%5E2%2Bxy%29%2B%28xy%2By%5E2%29 Group like terms


x%28x%2By%29%2By%28x%2By%29 Factor out the GCF of x out of the first group. Factor out the GCF of y out of the second group


%28x%2By%29%28x%2By%29 Since we have a common term of x%2By, we can combine like terms

So x%5E2%2Bxy%2Bxy%2By%5E2 factors to %28x%2By%29%28x%2By%29


So this also means that x%5E2%2B2xy%2By%5E2 factors to %28x%2By%29%28x%2By%29 (since x%5E2%2B2xy%2By%5E2 is equivalent to x%5E2%2Bxy%2Bxy%2By%5E2)


note: %28x%2By%29%28x%2By%29 is equivalent to %28x%2By%29%5E2 since the term x%2By occurs twice. So x%5E2%2B2xy%2By%5E2 also factors to %28x%2By%29%5E2



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So our expression goes from xy%28x%5E2%2B2xy%2By%5E2%29 and factors further to xy%28x%2By%29%5E2


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Answer:

So x%5E3y%2B2x%5E2y%5E2%2Bxy%5E3 completely factors to xy%28x%2By%29%5E2





# 2


32x%5E2y-2y%5E3 Start with the given expression


2y%2816x%5E2-y%5E2%29 Factor out the GCF 2y


2y%28%284x%29%5E2-%28y%29%5E2%29 Rewrite 16x%5E2 as %284x%29%5E2


2y%284x%2By%29%284x-y%29 Factor %284x%29%5E2-%28y%29%5E2 to get %284x%2By%29%284x-y%29 (use the difference of squares)



So 32x%5E2y-2y%5E3 completely factors to 2y%284x%2By%29%284x-y%29