# division of polynominals: (y^3 - 11y^2 + 6) / (y^2 - 3)

#### sara

##### New member
(y^3 - 11y^2 + 6) / (y^2 - 3)

when I divide the first term by the first term y^3/y^2 I get y
then multiple y above by the divisor (y^2-3) I get y^3-3y how can i subtract 11y^2 - 3y they are two different degree
help have I done something wronge in my math

Thank you

#### stapel

##### Super Moderator
Staff member
sara said:
(y^3 - 11y^2 + 6) / (y^2 - 3)

when I divide the first term by the first term y^3/y^2 I get y
then multiple y above by the divisor (y^2-3) I get y^3-3y how can i subtract 11y^2 - 3y they are two different degree
I'm not sure I follow...?

You have the following exercise:

Code:
        ---------------------
y^2 - 3 )y^3 - 11y^2 + 0y + 6
To get y[sup:1uz7qpqg]3[/sup:1uz7qpqg], the first term in the dividend, you have to multiply y[sup:1uz7qpqg]2[/sup:1uz7qpqg], the first term of the divisor, by y, so "y" goes on top:

Code:
          y
---------------------
y^2 - 3 )y^3 - 11y^2 + 0y + 6
Multiplying the divisor by y, you get y(y[sup:1uz7qpqg]2[/sup:1uz7qpqg] - 3) = y[sup:1uz7qpqg]3[/sup:1uz7qpqg] - 3y. This goes underneath, and is subtracted off:

Code:
y
---------------------
y^2 - 3 )y^3 - 11y^2 + 0y + 6
y^3         - 3y
---------------------
- 11y^2 + 3y + 6
But I'm afraid I don't understand what you did next...? Please reply with clarification. Thank you!

Eliz.

#### sara

##### New member
y

y^2-3 ) y^3 – 11y^2 + 0y + 6

y^3 – 3y this is were I was stuck. I did not know I could shift 3y, and subtract from 0y

So

y-11

y^2-3 ) y^3 -11y^2 + 0y + 6

-y^3 + 3y

- 11y^2 + 3y

11y^2 - 33y is this right so far

30 y + 6

Thank you

#### kasie-tutor

##### Banned
Dear sara,

This is a good question. It will most definitely come up on a test (the kind of problem where the powers of the exponents don't go down linearly).

Whenever you are missing a term in your dividend (the part inside the divide symbol), so that the powers of your terms are not consecutive, you should add that term with a zero coefficient (like we did by adding the 0y).

Sometimes it is also helpful to do the same to the divisor (the part on the outside of the divide symbol that you are dividing by). Here I have added a 0y as well.

I also recommend lining up all the like terms in the whole problem in columns.

Code:
                             y - 11   <--  Quotient
______________________
y^2 + 0y -3 ) y^3 - 11y^2 + 0y + 6
_  y^3 +  0y^2 - 3y        <--  Subtract (Change signs
________________             of terms this row; add)

- 11y^2 + 3y + 6    <--  Bring down + 6
_   - 11y^2 + 0y + 33   <--  Distribute -11 over y^2-3
_________________
3y - 27   <--  Remainder