10x^4 -27x³ -110x² - 27x +10 = 0
⇔10x³( x+2) -47 x² ( x+2)-16x ( x+2)+5( x+2) =0
⇔( x+2) ( 10x³ - 47x² -16x + 5) =0
⇔( x+2) [ 10x² ( x-5) +3x( x-5) -( x-5)] =0
⇔ ( x+2 ) (x -5) ( 10x² +3x -1) =0
⇔\(\left[ \begin{array}{l}x =-2\\x =5 \\10x^{2}+3x -1 =0 \end{array} \right.\)
Xét 10x² +3x -1=0
Δ = b²-4ac = 3² - 4 . 10 . (-1) = 49
⇒pt có 2 nghiệm phân biệt
$x_{1}$ = $\frac{-3+ \sqrt[]{49} }{20}$ = $\frac{1}{5}$
$x_{1}$ = $\frac{-3- \sqrt[]{49} }{20}$ = $\frac{-1}{2}$
Vậy x∈{ -2 ; 5 ; 1/5 ; -1/2 }
