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!
$\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!