x(do potegi 3)+2x(do potegi 2)-16x-32=0
pomocy :(
x(do potegi 3)+2x(do potegi 2)-16x-32=0x²(x+2)-16(x+2)(x+2)(x²-16)(x+2)(x-4)(x+4)Rozwiązanie:x∈ -4,-2,4
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x(do potegi 3)+2x(do potegi 2)-16x-32=0
x²(x+2)-16(x+2)
(x+2)(x²-16)
(x+2)(x-4)(x+4)
Rozwiązanie:
x∈ -4,-2,4