'Rooty' polynomials!(part 2)

The division algorithm gives us that $f(x) = g(x)q(x) + r(x)$ .

Since $q(x) = r(x)$ , we have that $f(x) = g(x)r(x) + r(x) \longrightarrow f(x) = r(x)(g(x) + 1)$ .

The Rational Root Theorem helps us determine that $-1$ is a root of $f(x)$ . Thus, by the Remainder Factor Theorem , $x + 1$ is a factor of $f(x)$ , which can thus factor into $f(x) = (x + 1)(x^2 + x + 2).$

We have two representations for $f(x)$ , namely $f(x) = r(x)(g(x) + 1$ and $f(x) = (x + 1)(x^2 + x + 2).$ Since, $\text{deg}(g(x)) \geq \text{deg}(r(x))$ , it must be that $g(x) = x^2 + x + 1.$

Hence , $g(5) = 5^2 + 5 + 1 = 25 + 5 + 1 = \huge\boxed{31}$

We note that:

$f(x) = q(x)g(x)+r(x) = q(x)(g(x) +q(x) = q(x)(g(x)+1)$

$\Rightarrow \dfrac {f(x)}{q(x)} = g(x) + 1$

We also note that:

$f(x) = x^3+2x^2+3x+2=(x+1)(x^2+x+2)$

Assuming that $q(x)=x+1$ as it is lower degree than $x^2+x+2$ , then

$\Rightarrow \dfrac {f(x)}{x+1} = x^2+x+2 = g(x) +1$

$\Rightarrow g(x)=x^2+x+1 \quad \Rightarrow g(5) = 25+5+1 =\boxed{31}$

I think that first, we should check that r(x) is not a constant term. Which is quite easy but necessary to prove. First, assume that r(x) is n,n belongs to integers, then we find that 1/n is also an integer which forces n=1 that is a contradiction or n=-1.But then assuming that you meant non-negative integers, then n!=-1. The next steps are already written, therefore I am not writing them again.

We have f(x)=q(x){g(x)+1} Now, x^3+2x^2+3x+2 =(x+1)^3-(x^2-1) =(x+1){(x^2+x+1)+1} Thus, by inspection; g(x) =x^2+x+1 And g(5) =31(Answer)

$x^3 + 2x^2 + 3x + 2 = (x^2 + x +2)(x +1)$

By Division Algorithm $f(x) = q(x)g(x) + r(x)$ . Since $q(x)= r(x)$ ,

thus $f(x) = q(x)g(x) + q(x)$ . This gives us $f(x) = q(x)(g(x) + 1)$ .

By inspection, $g(x)+1 = x^2 + x +2$ , thus

$g(x) = x^2 + x + 1$ . Hence $g(5) = 31$