Question

$( x+3)^{3}-16x-48=(x-1)(x+3)(x+7)$
aici este o regula sau ceva daca ati putea sa mi-o spuneti va rog dau coronita

Answer 1

x³+3·x²·3+3·x·3²+3³-16x-48=(x²+3x-x-3)(x+7)
x³+9x²+27x+27-16x-48=(x²+2x-3)(x+7)
x³+9x²+11x-21=x³+7x²+2x²+14x-3x-21
x³+9x²+11x-21=x³+9x²+11x-21
x³-x³+9x²-9x²+11x-11x-21+21=0
x∉Ф

(x+3)³-16(x+3)=(x-1)(x+3)(x+7)
(x+3)[(x+3)²-16]=(x-1)(x+3)(x+7)                     : (x+3)
(x+3)²-16=(x-1)(x+7)
x²+6x+9-16=x²+7x-x-7
x²+6x-7=x²+6x-7
x²-x²+6x-6x-7+7=0
x∈Ф