Rational polynomial

Using Sum of Two Cubes in Part Angle Fraction and Part Low Fraction Putting $x^3$ Evidence , So Have :

$\frac{(x+2)(x^2-2x+4)}{x^3(x^2-2x+4)}$

Simplified , We So Have :

$\frac{f(x)}{g(x)} = \frac{x+2}{x^3} \Rightarrow \frac{f(4)}{g(4)} \Rightarrow \frac{6}{64} = \frac{3}{32}$

$f(x) + g(x) = x+2+x^3 \Rightarrow f(4) + g(4) = 4^3 + 4 +2 = 70$

By factorizing,

$\frac{f(x)}{g(x)}$ = $\frac{(x+2)(x^{2}-2x+4}{x^{3}(x^{2}-2x+4)}$

Since, gcd(f(x), g(x)) is 1, we remove the common term $(x^{2}-2x+4)$ from both the numerator and denominator. So we get

$\frac{f(x)}{g(x)}$ = $\frac{(x+2)}{x^{3}}$

Putting x=4,

$\frac{f(x)}{g(x)}$ = $\frac{6}{64}$

therefore, f(x)+g(x)=64 + 6= $\boxed{70}$

Step 1 Understand the problem and the conditions

When I first saw this question, I was quite puzzled over the definition of the notation $gcd(f(x),g(x)) =1$ . With the help of some friends (special credits to Sreejato Bhattacharya and Tan Gian Yion), I understood that notation as $f(x)$ and $g(x)$ haven't no common root or algebraic factors.

Second, note that the equation $\displaystyle \frac{f(x)}{g(x)} = \frac{x^3 +8}{x^5-2x^4+4x^3}$ may not be expressed in the simplest form. However, it ought to be if we want to find the exact value of each of the functions when the specific value of $x$ is substitute in, since the question has the condition $gcd(f(x),g(x)) =1$ .

* Step 2 * Simplifying the expression

Note that the expression simplifies into: $\frac{x^3 +8}{x^5-2x^4+4x^3} = \frac{x^3 +8}{x^3(x^2-2x+4)}$

However, at this stage, the expression has yet to be in its simplest form as we desire. One powerful yet simple tool that comes to mind when we face an algebraic problem requiring factorisation is the long division method. Here, we can use it to simplify our expression as follows:

$\frac{x^3 +8}{x^3(x^2-2x+4)} = \frac{(x+2)(x^2-2x+4)}{x^3(x^2-2x+4)} = \frac{(x+2)}{x^3}$ .

Here, we have arrived at our simplest form of the equation. Therefore, substituting $x=4$ , we get $f(4) = 6$ and $g(4)=4^3 = 64$ . Thus, $f(4)+g(4) = \boxed{70}$ .

$\frac{x^3 + 8}{x^5 - 2x^4 + 4x^3} = \frac{(x+2)(x^2 - 2x^2 + 4)}{x^3(x^2 - 2x^2 + 4)} = \frac{x+2}{x^3}$

Let $f(x) = (x-\alpha_1)(x-\alpha_2)...(x-\alpha_m)$ and $g(x) = (x-\beta_1)(x-\beta_2)...(x-\beta_n)$

Since $gcd(f(x), g(x)) = 1$ , the sets $\{\alpha_1, \alpha_2, \dots, \alpha_m\}$ and $\{\beta_1, \beta_2, \dots, \beta_n\}$ are mutually exclusive.

$f(x) = x+2$ and $g(x) = x^3$ satisfy these conditions. Therefore, $f(4) + g(4) = 4 + 2 + 4^3 = \boxed{70}$

$\dfrac{f(x)}{g(x)}=\dfrac{x^3+8}{x^5-2x^4+4x^3}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^3\left(x^2-2x+4\right)}=\dfrac{x+2}{x^3}$

$\gcd\left(f(x),g(x)\right)=1=\gcd\left((x+2),x^3\right)$

$\therefore ~~~ f(x)=x+4 ~~~~~~~~~~~~ g(x)=x^3 ~~~~~~[\text{Since they are monic}]$

$\therefore ~~~ f(4)+g(4)=\fbox{70}$

Uma vez que $\text{mdc} \left ( f(x), g(x) \right ) = 1$ , temos:

$\\ \frac{f(x)}{g(x)} = \frac{x^3 + 8}{x^5 - 2x^4 + 4x^3} \\\\\\ \frac{f(x)}{g(x)} = \frac{(x + 2)(x^2 - 2x + 4)}{x^3(x^2 - 2x + 4)} \\\\\\ \frac{f(x)}{g(x)} = \frac{(x + 2)}{x^3} \\\\\\ \begin{cases} f(x) = x + 2 \Rightarrow f(4) = 4 + 2 \Rightarrow \boxed{f(4) = 6} \\ g(x) = x^3 \Rightarrow g(4) = 4^3 \Rightarrow \boxed{g(4) = 64} \end{cases}$

Por conseguinte,

$\\ f(4) + g(4) = 6 + 64 \\\\ \boxed{\boxed{f(4) + g(4) = 70}}$

take $x^{3}$ as a common factor from the denimenator you will get $x^{3}(x^{2}-2x+4)$ you should observe ,somehow, that $x^{3}+8$ can be factorized to $(x^{2}-2x+4)(x+2)$ $x^{2}-2x=4$ is cancelled from the denomenator and nometnator to get $\frac{x+2}{x^{3}}$ you can take the top as f(x) and the bottom as g(x) now the do the addition to get $\boxed{70}$

Note that $x^3+8=(x+2)(x^2-2x+4)$ and $x^5-2x^4+4x^3=x^3(x^2-2x+4)$ , which reduces the given fraction to $\frac{f(x)}{g(x)}=\frac{x+2}{x^3}$ . As these two terms have no common factor at all, we know that $f(x)=x+2$ and $g(x)=x^3$ , so $f(4)+g(4)=(4+2)+(4^3)=6+64=\boxed{70}$ .

Note that since the $\gcd(f(x),g(x))=1$ , then, they're co-prime functions. So, therefore, if we simplify $\frac{f(x)}{g(x)}$ To its lowest terms, we can easily recover both $f(x)$ and $g(x)$ by looking at the numerator and denominator, respectively.

Note: for a more technical reason, it is because $\mathbb{R}[x]$ has only a single unique representation of any polynomial apart from trivial cases--that is, other representations which have coefficients that are 0.

We find that we can factor out from both the sum of cubes (which factors into $x^3+8=(x+2)(x^2-2x+4)$ and the other case, from which we can clearly factor out an $x^3$ term. Dividing through, gives us: $\frac{f(x)}{g(x)}=\frac{x+2}{x^3}$ Then, $f(x)=x+2$ and $g(x)=x^3$ , for which, after we plug in 4, we get: $f(4)+g(4)=4+2+4^3=70$

The given fraction can be expressed as follows.

$\frac{x^{3}+8}{x^{5}-2x^{4}+4x^{3}}$

= $\frac{(x+2)(x^{2}-2x+4)}{x^{3}(x^{2}-2x+4)}$

= $\frac{x+2}{x^{3}}$

Therefore, we get $f(x)$ $=$ $x+2$ , $g(x)$ $=$ $x^{3}$ .

Plug in those numbers, we get $4+2+4^{3}$ $=$ $\boxed{70}$

Seeing the expression on the R.H.S, the numerator factors off into (x+2) ( x^2 -2x+4) and the denominator factors off into x^3 * (x^2-2x+4)...................so the fraction, on simplification, becomes (x+2)/(x^3). Uptil now, friends, it was just school algebra, now is the main essence of the problem. Here, it's given that gcd(f,g)=1.......so f/g will be the quotient of 2 pairwise prime polynomials..............and as x+2 and x^3 are pairwise polynomials, so f=x+2 , g=x^3, giving f(4)+g(4)=70.