Question

2. Descompuneti in factori:
a) y la a doua-3y+2;
b) a la a doua y la a doua-5a la a doua y+6a la a doua;
c) x la a doua y-yx-2x+2;
d) a la a doua+a la a doua b-3a-3ab;
e) yx la a doua-6xy+16y;
f) yx la a doua+5xy+4y;
g) x la a patra-11x la a doua+30.

Answer 1

a)
 y²-3y+2=0
Δ=b²-4ac=9-4×2=1
x₁=(-b+√Δ)/2a=(3+1)/2=2
x₂=(-b-√Δ)/2a=(3-1)/2=1
a(x-x1)(x-x₂)=(x-2)(x-1)

b)
a²y²-5a²y+6a²=0
Δ=b²-4ac=25a⁴-24a²×a²=25a⁴-24a⁴=a⁴
x₁=(-b+√Δ)/2a=(5a²+a²)/2a²=3
x₂=(-b-√Δ)/2a=(5a²-a²)/2a²=2
a(x-x1)(x-x₂)=(x-2)(x-3)

c)
x²y-yx-2x+2=xy(x-1)+2(x-1)=(x-1)(xy+2)

d)
a²+a²b-3a-3ab=a²(1+b)-3a(1+b)=(1+b)(a²-3a)

e) yx la a doua-6xy+16y;
yx²-6xy+16y=y(x²-6x+16)
x²-6x+16=0
Δ=b²-4ac=36-4×16=36-64=-28
x₁=(-b+√Δ)/2a=(6+√(-28))/2=6+2i√7
x₂=(-b-√Δ)/2a=(6+√(-28))/2=6-2i√7
a(x-x1)(x-x₂)=(x-6-2i√7)(x-6+2i√7)


f)
 yx la a doua+5xy+4y;
yx²+5xy+4y=y(x²+5x+4)
(x²+5x+4)=0
Δ=b²-4ac=25-4×4=9
x₁=(-b+√Δ)/2a=(-5+3)/2=1
x₂=(-b-√Δ)/2a=(-5-1)/2=-3
a(x-x1)(x-x₂)=(x-1)(x+3)