Question

Sa se descompună in factori:
a)x la patrat+5x+6
x la patrat+7x+12
x la patrat+7x+10
x la patrat+9x+20
b)x la patrat-5x+6
x la patrat-7x+12
x la patrat-7x+10
x la patrat-49y+20
c)x la patrat+8x+15
x la patrat+9x+14
x la patrat+11x+30
x la patrat+15x+56

Answer 1

a)x²+5x+6=
se descompune astfel ax²+bx+c=0 devine a(x-x1)(x-x₂) unde x₁ si x₂ sunt solutiei ecuatiei de grad 2.
x²+5x+6==0
Δ=b²-4ac=25-4×6=25-24=1
x₁=(-b+√Δ)/2a=(-5+1)/2=-2
x₂=(-b-√Δ)/2a=(-5-1)/2=-3
a(x-x1)(x-x₂)=(x+2)(x+3)
verificare:(x+2)(x+3)=x²+3x+2x+6=x²+5x+6
 
x²+7x+12
Δ=b²-4ac=49-4×12=49-48=1
x₁=(-b+√Δ)/2a=(-7+1)/2=-3
x₂=(-b-√Δ)/2a=(-7-1)/2=-4
a(x-x1)(x-x₂)=(x+3)(x+4)
verificare:(x+3)(x+4)=x²+3x+4x+12=x²+7x+12


x²+7x+10=0
Δ=49-40=9
x₁=(-b+√Δ)/2a=(-7+3)/2=-2
x₂=(-b-√Δ)/2a=(-7-3)/2=-5
a(x-x1)(x-x₂)=(x+2)(x+5)


x²+9x+20
Δ=81-80=1
x₁=(-b+√Δ)/2a=(-9+1)/2=-4
x₂=(-b-√Δ)/2a=(-9-1)/2=-5
a(x-x1)(x-x₂)=(x+4)(x+5)


b)x²-5x+6
Δ=25-24=1
x₁=(-b+√Δ)/2a=(5+1)/2=3
x₂=(-b-√Δ)/2a=(5-1)/2=2
a(x-x1)(x-x₂)=(x-4)(x-2)


x²-7x+12
Δ=49-48=1
x₁=(-b+√Δ)/2a=(7+1)/2=4
x₂=(-b-√Δ)/2a=(7-1)/2=3
a(x-x1)(x-x₂)=(x-4)(x-3)

x²-7x+10
Δ=49-40=9
x₁=(-b+√Δ)/2a=(7+3)/2=5
x₂=(-b-√Δ)/2a=(7-3)/2=2
a(x-x1)(x-x₂)=(x-5)(x-2)


c)x ²+8x+15
Δ=64-60=4
x₁=(-b+√Δ)/2a=(-8+2)/2=-3
x₂=(-b-√Δ)/2a=(-8-2)/2=-5
a(x-x1)(x-x₂)=(x+3)(x+5)

x ²+9x+14
Δ=81-56=25
x₁=(-b+√Δ)/2a=(-9+5)/2=-2
x₂=(-b-√Δ)/2a=(-9-3)/2=-7
a(x-x1)(x-x₂)=(x+2)(x+7)

x ²+11x+30
Δ=121-120=1
x₁=(-b+√Δ)/2a=(-11+1)/2=-6
x₂=(-b-√Δ)/2a=(-11-1)/2=-5
a(x-x1)(x-x₂)=(x+6)(x+5)


x²+15x+56 
Δ=225-224=1
x₁=(-b+√Δ)/2a=(-15+1)/2=-7
x₂=(-b-√Δ)/2a=(-15-1)/2=-8
a(x-x1)(x-x₂)=(x+7)(x-+8)