# Quick division questions

• Aug 27th 2009, 08:09 PM
clips
Quick 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 PM
apcalculus
\$\displaystyle
Quote:

Originally Posted by clips
can someone help me with this,its very quick
(3y^3 - 9y^2 - 3)

\$
Quote:

Originally Posted by clips
divided by \$\displaystyle (3y^2 +1)\$

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

\$\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 AM
Defunkt
Quote:

Originally Posted by clips
can someone help me with this,its very quick
(3y^3 - 9y^2 - 3) divided by (3y^2 +1)

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

(Thinking)