Few questions different topics

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In summary, the constants p,q, and r for the equation 2y^2 - 9y + 14 = p(y-1)(y-2) + q(y-1) + r are p=2, q=-3, and r=15. For the second part, the expression (y+1)^4 - y^4 < 4(y+1)^3 can be simplified by expanding and canceling out large powers of y.
  • #1
lionely
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Homework Statement



Find the constants p,q and r such that
2y2 - 9y+14 = p(y-1)(y-2)+q(y-1) +r

2) Deduce that (y+1)4 - y^4 < 4(y+1)^3





The attempt at a solution

Well I expanded the right side

and just tried comparing the two equations..

so I got p =2 ,r =10, q=-3

is this the correct way of doing this?

2) (y+1)4 - y^4 - 4(y+1)^3<0

I did this first, now I don't know where to carry it.
 
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  • #2
p and q are correct but r is wrong.

Are you really supposed to use these values for p q and r to deduce the second part? Because I don't see the connection at all
 
  • #3
NO the 2nd part of the question is a totally different question ,sorry. is r 14?
 
  • #4
No it's not 14 either. Why don't you write out how you solved for r and we can pinpoint the mistake being made.

For the second part once you've gotten to that point I recommend expanding everything and canceling a lot of large powers of y
 
  • #5
I expanded the expression and got py^2 - 3py -12p + qy -q +r

so I just compared it to the equation stated and said p is =2 because there's also a two on the y^2 in the first equation. Then I said since p = 2 the -3py is now -6y and in the equation there was a -9y so q needs to be -3 to get a -3y. the r ,-12p and the -q need to add up to the 14 so is r 35?
 
Last edited:
  • #6
That should be a +2p not a -12p
 
  • #7
ohh I see it, so r should be 15?!
 

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