Dividing polynomials

BabyBlue

New member
Joined
Jun 1, 2007
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15

Sometimes I get stuck and confused in polynomials!
Here is one that has me puzzled

2P(squared)+ 7P- 39 by 2p-7
I get puzzled when both the positive sevens are added, and it should be 14P, but they say the answer is 7P. :shock:
Because the divisor is negative 7, when you're taking away it should be positive. Right?
 

skeeter

Senior Member
Joined
Dec 15, 2005
Messages
2,396
Code:
        p   + 7
      ----------------------
2p-7 | 2p^2 + 7p - 39
       2p^2 - 7p
       -------------------
             14p - 39
             14p - 49
             ----------
                   10
 

BabyBlue

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Jun 1, 2007
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I just figured out where I was going wrong. Thanks, I feel stupid.
:oops:
 

jwpaine

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Mar 10, 2007
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BabyBlue said:
I just figured out where I was going wrong. Thanks, I feel stupid.
:oops:
Don't ever feel stupid. Life is about learning, and exploring. If you did everything perfect and knew everything, than there would be nothing left in life.
 

jonboy

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Joined
Jun 8, 2006
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547
jwpaine said:
BabyBlue said:
I just figured out where I was going wrong. Thanks, I feel stupid.
:oops:
Don't ever feel stupid. Life is about learning, and exploring. If you did everything perfect and knew everything, than there would be nothing left in life.
Couldn't have said it better :!:
 

BabyBlue

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Joined
Jun 1, 2007
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15
Hello,
I just keep having problems arranging the polynomials. Like for instance this problem:
b^9+6b^6+b^4+9b^3+4b+8 by b^3+4

I just think hmm.. lets it logical to start with the ninth power and then so on, but it is not working and matching up. Do you get what I mean by matching up? Like the second term of the divisor doesn't divide evenly into it. The powers are different, and they should not be. Corrrect?

I am home schooled so it is kind of hard.

Thanks,
Amy
:)
 

skeeter

Senior Member
Joined
Dec 15, 2005
Messages
2,396
BabyBlue said:
Hello,
I just keep having problems arranging the polynomials. Like for instance this problem:
b^9+6b^6+b^4+9b^3+4b+8 by b^3+4

I just think hmm.. lets it logical to start with the ninth power and then so on, but it is not working and matching up. Do you get what I mean by matching up? no. Like the second term of the divisor doesn't divide evenly into it. it doesn't have to ... the lead term is driving the division process. The powers are different, and they should not be. Corrrect?

I am home schooled so it is kind of hard.

Thanks,
Amy
:)
I'm not going to post all my work, but I can tell that the process is just straight long division and the quotient is ...

b<sup>6</sup> + 2b<sup>3</sup> + b + 1 with remainder 4/(b<sup>3</sup> + 4).

btw ... in future, start a new problem with a new post.
 

BabyBlue

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Jun 1, 2007
Messages
15
Skeeter,
next time I'll post a new one :wink:
I guess I just didn't know that they didn't have to have the same powers. I thought that if they did not that meant you were doing something wrong in the process. I'll work on it,
Amy :p
 
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