Question

Suma modulelor valorilor lui a,b∈R pentru care polinomul $f=X^4+aX^3+bX^2+X+5$ se divide cu $X^2 -X - 5$
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Answer 1

   
[tex]\displaystyle\\ f=x^4+ax^3+bx^2+x+5 \\ b=m+n\\ \Longrightarrow f=x^4+ax^3+bx^2+x+5=x^4+ax^3+mx^2+nx^2+x+5\\\\ \frac{f}{x^2-x-5}=\frac{x^4+ax^3+mx^2+nx^2+x+5}{x^2-x-5}=\\\\ =\frac{(x^4+ax^3+mx^2)+(nx^2+x+5)}{x^2-x-5}=\\\\ = \frac{x^4+ax^3+mx^2}{x^2-x-5}+\frac{nx^2+x+5}{x^2-x- 5}=\\\\ = \frac{x^2(x^2+ax+m)}{x^2-x-5}+\frac{-1(-nx^2-x-5)}{x^2-x-5}\\\\ x^2+ax+m=x^2-x-5\Longrightarrow a=-1~\text{si}~m=-5\\ -nx^2-x-5=x^2-x-5\Longrightarrow n=-1\\ b=m+n=-5-1=-6\\ \text{Solutia: } \boxed{a=-1~\text{si}~b=-6}[/tex]

$\Longrightarrow~~f=x^4+ax^3+bx^2+x+5 = {\bf x^4-x^3-6x^2+x+5 }$

S = |a| + |b| = |-1| + |-6| = 1 + 6 = 7

Answer 2

[tex]\it f= X^4+aX^3+bX+X+5 \\ \\ X^2-X-5|f \Rightarrow f = (X^2-X-5)(X^2+pX+q) \ \ \ \ \ (1) \\ \\ (X^2-X-5)(X^2+pX+q) = \\ \\ = X^4+pX^3+qX^2-X^3-pX^2-qX-5q = \\ \\ = X^4+(p-1)X^3+(q-p-5)X^2+(-q-5p)X-5q \ \ \ \ (2) [/tex]


[tex]\it (1), (2) \Rightarrow \begin{cases} \it -5q=5 \Rightarrow q=-1 \ \ \ \ (3) \\ \\ \it -q-5p=1 \stackrel{(3)}{\Longrightarrow } -1-5p=-1 \Rightarrow p=0\ \ \ \ (4) \\ \\ \it q-p-5=b \stackrel{(3),(4)}{\Longrightarrow} -1-0-5=b \Rightarrow b = -6 \\ \\ \it p-1= a \stackrel{(4)}{\Longrightarrow} a = -1\end{cases}[/tex]


$\it |a| +|b| = |-1| + |-6| = 1+6=7$