Verified Answer :

sakethmalyala

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:

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

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 \(9x^3-11x-7x\)

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