Question

Descompune asociind convenabil si scotand factor comun:

Exemplu:x^3+x^2y+xy^2+y^3=x^2(x+y)+y^2(x+y)=(x+y)(x^2+y^2)

a)a^3+a^2b+ab^2+b^3
b)m^3-3m^2+3m-9
c)x^2-2ax+xy-2ay
d)3x+x^2+3y-y^2
e)xz+xt+yz+yt
f)p^5-p^4+3p^3-3p^2
g)a^2-b^2+xa+xb
h)a^2+2a-b^2-2b

Va rog frumos!

Answer 1

Rezolvarea e in poza.Sper ca te-am ajutat :)

Answer 2

a) a^3 + a^2*b + a*b^2 + b^3 = a^2(a+b) + b^2(a+b) = (a+b)(a^2+b^2)

b) m^3 - 3m^2 + 3m - 9 = m^2(m-3) + 3(m-3) = (m-3)(m^2+3)

c) x^2 - 2ax + xy - 2ay = x(x-2a) + y(x-2a) = (x-2a)(x+y)

d) 3x + x^2 + 3y - y^2 = 3x + 3y + x^2 - y^2 = 3(x+y) + (x+y)(x-y) = (x+y)(3+x-y)

e) xz + xt + yz + yt = xz + yz + xt + yt = z(x+y) + t(x+y) = (x+y)(z+t)

f) p^5 - p^4 + 3p^3 - 3p^2 = p^4(p-1) + 3p^2(p-1) = (p-1)(p^4 + 3p^2) = (p-1)p^2(p^2+3)

g) a^2 - b^2 + xa + xb = (a+b)(a-b) + x(a+b) = (a+b)(a-b+x)

h) a^2 + 2a - b^2 - 2b = a^2 - b^2 + 2a - 2b = (a+b)(a-b) + 2(a-b) = (a-b)(a+b+2)