SOLUTION: "Completely factor each polynomial. If the polynomial cannot be factored, state so." 25y^3-10y^2+y y(25y^2-10y+1) I know this is a perfect square polynomial, but I'm not s

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: "Completely factor each polynomial. If the polynomial cannot be factored, state so." 25y^3-10y^2+y y(25y^2-10y+1) I know this is a perfect square polynomial, but I'm not s      Log On


   

Question 880974: "Completely factor each polynomial. If the polynomial cannot be factored, state so."
25y^3-10y^2+y
y(25y^2-10y+1)
I know this is a perfect square polynomial, but I'm not sure what to do after this. Thanks!!!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let's factor 25y%5E2-10y%2B1


Looking at the expression 25y%5E2-10y%2B1, we can see that the first coefficient is 25, the second coefficient is -10, and the last term is 1.


Now multiply the first coefficient 25 by the last term 1 to get %2825%29%281%29=25.


Now the question is: what two whole numbers multiply to 25 (the previous product) and add to the second coefficient -10?


To find these two numbers, we need to list all of the factors of 25 (the previous product).


Factors of 25:
1,5,25
-1,-5,-25


Note: list the negative of each factor. This will allow us to find all possible combinations.


These factors pair up and multiply to 25.
1*25 = 25
5*5 = 25
(-1)*(-25) = 25
(-5)*(-5) = 25

Now let's add up each pair of factors to see if one pair adds to the middle coefficient -10:


First NumberSecond NumberSum
1251+25=26
555+5=10
-1-25-1+(-25)=-26
-5-5-5+(-5)=-10



From the table, we can see that the two numbers -5 and -5 add to -10 (the middle coefficient).


So the two numbers -5 and -5 both multiply to 25 and add to -10


Now replace the middle term -10y with -5y-5y. Remember, -5 and -5 add to -10. So this shows us that -5y-5y=-10y.


25y%5E2%2Bhighlight%28-5y-5y%29%2B1 Replace the second term -10y with -5y-5y.


%2825y%5E2-5y%29%2B%28-5y%2B1%29 Group the terms into two pairs.


5y%285y-1%29%2B%28-5y%2B1%29 Factor out the GCF 5y from the first group.


5y%285y-1%29-1%285y-1%29 Factor out 1 from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.


%285y-1%29%285y-1%29 Combine like terms. Or factor out the common term 5y-1


%285y-1%29%5E2 Condense the terms.


So 25y%5E2-10y%2B1 factors to %285y-1%29%5E2.


In other words, 25y%5E2-10y%2B1=%285y-1%29%5E2.


Note: you can check the answer by expanding %285y-1%29%5E2 to get 25y%5E2-10y%2B1 or by graphing the original expression and the answer (the two graphs should be identical).


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So y%2825y%5E2-10y%2B1%29+ turns into y%285y-1%29%5E2+

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Final Answer: 25y%5E3-10y%5E2%2By+ completely factors to y%285y-1%29%5E2