Question

Let f(x) = 9x3+ 21x2^ - 14 and g(x) = 3x + 1. Find "f (x) over g (x)​

Answer 1

Answer:

$\frac{f(x)}{g(x)}=3x^2+6x-2-\frac{12}{3x+1}$

Step-by-step explanation:

The first function is $f(x)=9x^3+21x^2-14$

The second function is $g(x)=3x+1$.

$\frac{f(x)}{g(x)}=\frac{9x^3+21x^2-14}{3x+1}$

We perform the long division as shown in the attachment to obtain the quotient as: $Q(x)=3x^2+6x-2$ and remainder $R=-12$.

Therefore:

$\frac{f(x)}{g(x)}=3x^2+6x-2-\frac{12}{3x+1}$

where $x\ne -\frac{1}{3}$