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stew
11-01-2005, 10:08 PM
I'm having trouble with this problem. I'm supposed to factor 5y^5-5y^4-20y^3+20y^2. I come up with 5y^2(y-1)(y+2)(y-2). The book gives an answer of (y-1)(y+1)(y-2). How did they come up with this answer? I have tried everything. Can someone help?

soroban
11-01-2005, 10:44 PM
Hello, stew!

Be more assertive . . . you're right!

Factor 5y^5\,-\,5y^4\,-\,20y^3\,+\,20y^2

I come up with 5y^2(y-1)(y+2)(y-2)

The book gives an answer of (y-1)(y+1)(y-2)
First of all, there is no way to disregard the 5y^2

The book has typos . . . obviously!

We have: .5y^2[y^3\,-\,y^2\,-\,4y\,+\,4]

. . . = \;5y^2[y^2(y - 1) - 4(y - 1)]

. . . = \;5y^2(y - 1)[y^2 - 4]

. . . = \;5y^2(y - 1)(y + 2)(y - 2) . . . your answer!

Denis
11-01-2005, 10:45 PM
Guess what...you're correct, and the book is wrong!

You can check if you're correct by giving y any value,
and substituting this value in original equation, then in your solution:
they must equal each other.

Try y = 3:
original: 5y^5-5y^4-20y^3+20y^2 = 5(3^5)-5(3^4)-20(3^3)+20(3^2) = 450
your solution: 5y^2(y-1)(y+2)(y-2) = 5(3^2)(3-1)(3+2)(3-2) = 450

stew
11-02-2005, 09:02 AM
Thanks for the help guys. I had a sneaking suspicion that it was wrong, but how often does that happen? I knew you guys would be able to help. Usually when the book has an answer that I think is impossible, after I look at it for a while it turns out to be right. Thanks for the info!