Divide the following: (3x^3=4x-1)/(x^2+1) adn write the result in the form f(x)=d(x)*q(x)=r(x)
Thank you so very much
Surely a couple of those equal signs should be +, no?
So let me assume the question is $\displaystyle \frac{3x^3+4x-1}{x^2+1}$
$\displaystyle x^2$ goes into $\displaystyle 3x^3\quad$ $\displaystyle \quad 3x$ times (as $\displaystyle \frac{3x^3}{x^2}=3x$) so now $\displaystyle 3x^3+4x-1-(3x(x^2+1))=$
$\displaystyle =3x^3+4x-1-(3x^3+3x)$
$\displaystyle
=x-1
$
Since the degree of x-1 is 1, which is less than the degree of $\displaystyle x^2+1$ which is 2, we are done, and we have a remainder of x-1
So we get $\displaystyle 3x(x^2+1)+(x-1)$