Susan C. Posted December 5, 2011 Share Posted December 5, 2011 (edited) Dd did an even problem where there isn't an answer in the SM. I did the problem and want to make sure I did it correctly: (2y^5 - 5y^4 - 3y^2 - 6y - 23) divided by (y - 3) My answer came out to 2y^4 + y^3 + y^2 - 6 - 41/y-3 How did I do? Thanks! **Kiana is correct below** Edited December 5, 2011 by Susan C. Quote Link to comment Share on other sites More sharing options...
kiana Posted December 5, 2011 Share Posted December 5, 2011 Wolfram alpha differs with you. (scroll down to quotient and remainder, it will also show steps) http://www.wolframalpha.com/input/?i=%282y%5E5+-+5y%5E4+-+3y%5E2+-+6y+-+23%29+divided+by+%28y+-+3%29 They get 2y^4 + y^3 + 3y^2 + 6y + 12 remainder 13. Your error seems to be after the y^3 term. When you subtracted after finding y^3, you should've had (y^4 - 3y^2 - 6y - 23) - (y^4 - 3y^3), which is 3y^3 - 3y^2 - 6y - 23. The next term would then be 3y^2. It seems you accidentally ended up with y^3 leading. Perhaps you lost the 3 on the y^3? I can't give more details without seeing your working. Quote Link to comment Share on other sites More sharing options...
beaners Posted December 5, 2011 Share Posted December 5, 2011 Wolfram alpha differs with you. (scroll down to quotient and remainder, it will also show steps) http://www.wolframalpha.com/input/?i=%282y%5E5+-+5y%5E4+-+3y%5E2+-+6y+-+23%29+divided+by+%28y+-+3%29 They get 2y^4 + y^3 + 3y^2 + 6y + 12 remainder 13. Your error seems to be after the y^3 term. When you subtracted after finding y^3, you should've had (y^4 - 3y^2 - 6y - 23) - (y^4 - 3y^3), which is 3y^3 - 3y^2 - 6y - 23. The next term would then be 3y^2. It seems you accidentally ended up with y^3 leading. Perhaps you lost the 3 on the y^3? I can't give more details without seeing your working. I did long division and it looked like that. Is there any chance the actual problem also included -2y^3 ? By multiplying the answer in the OP and (y-3) I got the original polynomial plus -2y^3. But I was interrupted twice by someone who should have been asleep, so I may have made an error. Quote Link to comment Share on other sites More sharing options...
Susan C. Posted December 5, 2011 Author Share Posted December 5, 2011 Thank you, I was able to duplicate your answer. I'm very rusty..... here I am making her do synthetic division and I can barely do the regular division by hand, much less use the calculator.... Quote Link to comment Share on other sites More sharing options...
kiana Posted December 5, 2011 Share Posted December 5, 2011 I did long division and it looked like that. Is there any chance the actual problem also included -2y^3 ? By multiplying the answer in the OP and (y-3) I got the original polynomial plus -2y^3. But I was interrupted twice by someone who should have been asleep, so I may have made an error. Good point. No, you're correct. (If the problem was really (2y^5 - 5y^4 -2y^3 - 3y^2 - 6y - 23) divided by (y - 3), the answer was correct.) Quote Link to comment Share on other sites More sharing options...
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