can someone help me with this,its very quick

(3y^3 - 9y^2 - 3) divided by (3y^2 +1)

and if anyone can quickly explain the remainder theorem that would be appreciated. thanks a lot!

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- Aug 27th 2009, 08:09 PMclipsQuick division questions
can someone help me with this,its very quick

(3y^3 - 9y^2 - 3) divided by (3y^2 +1)

and if anyone can quickly explain the remainder theorem that would be appreciated. thanks a lot! - Aug 27th 2009, 08:49 PMapcalculus
$\displaystyle $

$\displaystyle 3y^2$ goes into $\displaystyle 3y^3 $ $\displaystyle y$ times so you have:

$\displaystyle y (3y^2 + 1) = 3y^3 + y$

Subtract the result from the dividend:

$\displaystyle 3y^3 - 9y^2 - 3 - (3y^3 + y) = -9y^2 - y - 3$

Now repeat the process:

$\displaystyle 3y^2$, the largest degree term of your divisor, goes into -$\displaystyle 9y^2$... -3 times so you have:

$\displaystyle -3 (3y^2+1) = -9y^2 - 3$

Subtract the result from $\displaystyle -9y^2 - y - 3$:

$\displaystyle -9y^2 - y - 3 - (-9y^2 - 3) = -y$

3y^2 is of greater degree than -y so -y is the remainder.

The quotient is y-3 and the remainder is -y

Check using:

Dividend = quotient * divisor + remainder

Remainder theorem:

If you divide a polynomial p(x) by a factor (x-a), the remainder is p(a). In plain English: the remainder is the y-value you get if you evaluate the polynomial at x=a.

example:

If you divide $\displaystyle f(x) = x^200 + 3x^99 + 2x - 1 $ by $\displaystyle (x-1)$, the remainder is $\displaystyle f(1)$, which is $\displaystyle 1 + 3 + 2 - 1 = 5$. Note that we can conclude about the remainder without knowing the quotient.

I hope this helps.

Good luck! - Aug 28th 2009, 02:03 AMDefunkt