Question

(3k-2)(2k^2+4k-7)=
(2+7q)(3q^3-4q^2+1)=
(a+5)(2a^4+3a^2-4=
(3+2m^3)(7m^5+3m^2-4)=
Приведите многочлен к стандартному виду

Answer 1

(3k-2)(2k²+4k-7) = 6k³+12k²-21k-4k²-8k+14= 6k³+8k²-29k+14
(2+7q)(3q³-4q²+1) = 6q³-8q²+2 +21q⁴-28q³+7q= 21q⁴-22q³-8q²+7q+2
(a+5)(2a⁴+3a²-4)=2a⁵+3a³-4a+10a⁴+15a²-20= 2a⁵+10a⁴+3a³+15a²-4a-20
(3+2m³)(7m⁵+3m²-4) = 21m⁵+9m²-12+14m⁸+6m⁵-8m³=
=14m⁸+27m⁵-8m³+9m²-12