Question

Se considera polinomul f= x³ - 2x² - 2x +1 .
Demonstrati ca (x2+x3)(x3+x1)(x1+x2)= -3, unde x1,x2,x3 sunt radacinile polinomului f.

Answer 1

[tex]f = $ x$^3-2$x$^2-2$x $+1 \\ \\ $x$_1+$x$_2+$x$_3 = -\dfrac{b}{a} \Rightarrow $ x$_1+$x$_2+$x$_3 = -\dfrac{-2}{1} \Rightarrow \\ \\ \Rightarrow $ x$_1+$x$_2+$x$_3 = 2 $ $ $ $($scoatem 3 relatii din relatia asta:) \\ \\ $\boxed{1}\quad $x$_2+$x$_3 = 2-$x$_1 \\ \\ \boxed{2} \quad$x$_3+$x$_1 = 2-$x$_2\\ \\ \boxed{3} \quad $x$_1+$x$_2 = 2-$x$_3 \\ \\ \\ [/tex]

$\Rightarrow ( x _2+ x _3)( x _3+ x _1)( x _1+ x _2) = (2- x _1)(2- x _2)(2- x _3) \\ \\ Polinomul x ^3-2 x ^2-2 x +1 se mai poate scrie sub forma de \\ factori ireductibili ca: ( x - x _1)( x - x _2)( x - x _3)$

$f = x ^3-2 x ^2-2 x +1 \\ f(2) = 2^3-2\cdot 2^2-2\cdot 2 + 1 = 8 -8-4+1 = -3 \\ \\ f = ( x - x _1)( x - x _2)( x - x _3)\\ f(2) = (2- x _1)(2- x _2)(2- x _3) \\ \\ Noi calcularam ca f(2) = -3 \Rightarrow (2- x _1)(2- x _2)(2- x _3) = -3$

$\Rightarrow \boxed{( x _2+ x _3)( x _3+ x _1)( x _1+ x _2) =-3}$

Answer 2

Fie S=x1+x2+x3

atunci expresia este
(S-x1)(S-x2)(S-x3)=f(S) conform formei ALGEBRICE a unui polinom cu coeficientul termenului dominant=1

iar S=x1+x2+x3=(viete)=-(-2)/1=2
atunci
f(2)=8-8-4+1=-3
 C.C.T.D.
 as  quicky as that!