Question

Let f(x) = 27x5 − 33x4 − 21x3 and g(x) = 3x2. Find f of x over g of x

Answer 1

I am assuming when you write 27x5 u mean "27 times x to the power of 5".

therefore, your answer is 9x3-11x2-7x

Answer 2

Answer:

$\frac{f(x)}{g(x)}=(9x^3-11x-7x$

Step-by-step explanation:

Given  function  $f(x) = 27x^5-33x^4-21x^3$ and $g(x)=3x^2$

We have to find f of x over g of x that is $\frac{f(x)}{g(x)}$

Consider ,

$f(x) = 27x^5-33x^4-21x^3$ and $g(x)=3x^2$

To find  $\frac{f(x)}{g(x)}$

Substitute the values for functions, we get,

$\frac{f(x)}{g(x)}=\frac{27x^5-33x^4-21x^3}{3x^2}$

Taking $3x^2$ common from numerator, we get,

$\frac{f(x)}{g(x)}=\frac{3x^2(9x^3-11x-7x)}{3x^2}$

Thus, $3x^2$  gets cancel, we get,

$\frac{f(x)}{g(x)}=9x^3-11x-7x$

Thus, value of function  f of x over g of x  is $9x^3-11x-7x$