Question

Descompuneți în factori următoarele expresii:

a) A = 3a² + 9c² + 6ab - 12ac - 6bc;
b) B = a² + 4ab + 4b² - a³ - 2a²b.​

Answer 1

 

$\displaystyle\bf\\a)\\\\A = 3a^2 + 9c^2 + 6ab - 12ac - 6bc=\\\\=3(a^2 + 3c^2 + 2ab - 4ac - 2bc)= (Reordonam)\\\\=3(a^2 + 2 a b - 4 a c - 2 b c + 3 c^2)=(Repartizam: -4ac=-3ac-ac)\\\\=3(a^2 + 2 a b - 3 a c - ac - 2 b c + 3 c^2)=(Grupam)\\\\=3[(a^2 + 2 a b - 3 a c)+ (- ac - 2 b c + 3 c^2)]=(Dam~factor~comun)\\\\=3[a(a + 2b - 3c)-c(a + 2b - 3c)]=(Dam~factor~comun~paranteza)\\\\=\boxed{\bf3(a + 2b - 3c)(a-c)}$

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$\displaystyle\bf\\b)\\\\B = a^2 + 4ab + 4b^2 - a^3 - 2a^2b=\\\\=-(-a^2 - 4ab - 4b^2 + a^3 + 2a^2b)=(Repartizam: - 4ab = -2ab -2ab )\\\\=-(-a^2 - 2ab - 2ab - 4b^2 + a^3 + 2a^2b)=(Reordonam)\\\\=-[a^3 - a^2 - 2ab + 2a^2b - 2ab - 4b^2]= (Grupam)\\\\=-[(a^3 - a^2 - 2ab) + (2a^2b - 2ab - 4b^2)]=(Dam~factor~comun)\\\\=-[a(a^2 - a - 2b) + 2b(a^2 - a - 2b)]=(Dam~factor~comun~paranteza)\\\\=\boxed{\bf-(a^2 - a - 2b)(a+2b)}$