Question

Descompuneți în factori :
a) 4x²+4x-35
b) 9x²+12x-45
c)4x²+20x-11
d)9x²+24x-33
Ajutor , nu le știu !!​

Answer 1

Răspuns:

) 4x²+4x-35 = (2x-5)(2x+7)=0

Δ=16+4×4×35=576

x₁=(-4+√576)/8=(-4+24)/8=20/8=5/2

x₁=(-4-√576)/8=(-4-24)/8=-28/8=-7/2

b) 9x²+12x-45 =3(3x²+4x-15)=3(3x-5)(x+3)=0

Δ=16+4×3×15=16+180=196

x₁=(-4+√196)/6=(-4+14)/6=10/6=5/3

x₁=(-4-√196)/6=(-4-14)/6=-18/6=-3

c) 4x²+20x-11=(2x-1)(2x+11)=0

Δ=400+4×4×11=400+176=576

x₁=(-20+√576)/8=(-20+24)/8=4/8=1/2

x₁=(-20-√576)/8=(-20-24)/8=-44/8=-11/2

d) 9x²+24x-33=3(3x²+8x-11)=3(x-1)(2x+11)=0

Δ=64+4×3×11=64+132=196

x₁=(-8+√196)/6=(-8+14)/6=6/6=1

x₁=(-8-√196)/6=(-8-14)/6=-22/6=-11/2

e) 3x²+7x+4 = (x+1)(3x+4)

Δ=49-4×3×4=49-48=1

x₁=(-7+√1)/6=(-7+1)/6=-6/6=-1

x₁=(-7-√1)/6=(-7-1)/6=-8/6=-4/3

f) 5x²+8x+3 = (x+1)(5x+3)

Δ=64-4×3×5=64-60=4

x₁=(-8+√4)/10=(-8+2)/10=-6/10=-3/5

x₁=(-8-√4)/10=(-8-2)/10=-1

a) 4x²+4x-35 = 4x²-10x+14x-35=2x(2x-5)+7(2x-5)=(2x-5)(2x+7)=0

b) 9x²+12x-45 =3(3x²+4x-15)=3(3x²-5x+9x-15)=3[x(3x-5)+3(3x-5)= 3(3x-5)(x+3)=0

c) 4x²+20x-11=4x²-2x+22x-11=2x(2x-1)+11(2x-1)=(2x-1)(2x+11)=0

d) 9x²+24x-33=3(3x²+8x-11)=3(2x²-2x+11x-11)=3[2x(x-1)+11(x-1]= 3(x-1)(2x+11)=0

e) 3x²+7x+4 =3x²+3x+4x+4=3x(x+1)+4(x+1)= (x+1)(3x+4)

f) 5x²+8x+3 =5x²+5x+3x+3=5x(x+1)+3(x+1)= (x+1)(5x+3)

Explicație pas cu pas: