Răspuns:
Explicație pas cu pas:
a) E(x)=x³+(x-3)²+2(x-2)(x+2)+8x-1
=x³+x²-6x+9+2(x²-4)+8x-1
= x³+x²-6x+9+2x²-8+8x-1
= x³+3x²+2x
=x(x²+3x+2)
=x(x²+x+2x+2)
= x[x(x+1)+2(x+1)]
=x(x+1)(x+2)
b) E(n)=n²+5n, n≠0, n∈Z
E(n)=n(n+1)(n+2) ⇒n(n+1)(n+2)=n²+5n ⇒n(n+1)(n+2)=n(n+5) Deoarece n≠0, putem imparti egalitatea cu n ⇒(n+1)(n+2)=n+5 ⇒n²+2n+n+2-n-5=0 ⇒
n²+2n-3=0 Ecuatie de gradul 2. Calculam Δ ⇒Δ=2²-4·1·(-3)=4+12=16 ⇒
n₁,₂=(-2±√Δ)/2 ⇒n₁=(-2-4)/2=-6/2=-3 ⇒n₁=-3
n₂=(-2+4)/2 =2/2=1 ⇒n₂=1
