Question

f=X³+mX²+nX+6 g=X²-X-2 determinați m si n astfel incat f sa fie divizibil cu g

Answer 1

$\displaystyle\\ f=x^3+mx^2+nx+6\\ g=x^2-x-2\\ f ~\vdots~ g \text{Calculam radacinile lui g}\\\\ x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{1\pm\sqrt{1+4\cdot2}}{2} =\frac{1\pm\sqrt{9}}{2}= \frac{1\pm3}{2}\\\\ x_1 = \frac{1-3}{2}=\boxed{-1}\\\\ x_2 = \frac{1+3}{2}=\boxed{2}$

$\displaystyle\\ \text{Daca f divizibil cu g atunci x1 si x2 sunt si radacini ale lui f.}\\ \text{Folosim relatia lui viete:} ~~~x_1x_2x_3 = \frac{-d}{a} \text{ pentru a afla pe x3}\\\\ x_1x_2x_3 = \frac{-d}{a}\\\\ x_1x_2x_3 = -6\\\\ x_3 = \frac{-6}{x_1x_2} = \frac{-6}{(-1)\cdot 2} = \frac{-6}{-2} =\boxed{3}\\\\ \text{Avem radacinile polinomului f(x):}\\ \boxed{x_1=-1;~x_2=2;~x_3=3}\\ \text{Reconstituim polinomul f(x)}\\\\ f(x) = (x-x_1)(x-x_2)(x-x_2)\\\\ f(x)=(x+1)(x-2)(x-3)$

$\displaystyle\\ f(x)=(x^2+x-2x-2)(x-3) =\\\\ = (x^2-x-2)(x-3) =\\\\ =x^3-x^2-2x-3x^2+3x+6=\\\\ =x^3-4x^2+x+6\\\\ f(x)=x^3-4x^2+x+6\\\\ f(x)=x^3+mx^2+nx+6\\\\ \Longrightarrow~~~\boxed{\bf m=-4}~~\bf si~~ \boxed{\bf n=1}$