Respuestas
La unión de cada trinomio con su respectiva factorización es:
a. 3a²+8a+5 = (a+1)(3a+5)
b. 13a²-7a-6 = (a-1)(13a+6)
c. 30a²+17a-21 = (5a-3)(6a+7)
d. 21a²+11a-2 = (7a-1)(3a+2)
e. 6a²+22a+20 = 2(3a+5)(a+2)
f. 8a²-2a-15 = (2a-3)(4a+5)
Explicación paso a paso:
a. 3a²+8a+5
Aplicar resolvente;
a₁,₂ = [-b±√(b²-4ac)]/2a
Sustituir;
a₁,₂ = [-8±√(8²-4(3)(5))]/2(3)
a₁,₂ = [-8±√4]/6
a₁ = -1
a₂ =-5/3
3a²+8a+5 = (a+1)(3a+5)
b. 13a²-7a-6
Aplicar resolvente;
a₁,₂ = [-b±√(b²-4ac)]/2a
Sustituir;
a₁,₂ = [7±√(7²-4(13)(-6))]/2(13)
a₁,₂ = [7±√(361)]/26
a₁,₂ = [7±19]/26
a₁ = 1
a₂ =-6/13
13a²-7a-6 = (a-1)(13a+6)
c. 30a²+17a-21
Aplicar resolvente;
a₁,₂ = [-b±√(b²-4ac)]/2a
Sustituir;
a₁,₂ = [-17±√(17²-4(30)(-21))]/2(30)
a₁,₂ = [-17±√(2809)]/60
a₁,₂ = [-17±53]/60
a₁ = 3/5
a₂ =-7/6
30a²+17a-21 = (5a-3)(6a+7)
d. 21a²+11a-2
Aplicar resolvente;
a₁,₂ = [-b±√(b²-4ac)]/2a
Sustituir;
a₁,₂ = [-11±√(11²-4(21)(-2))]/2(21)
a₁,₂ = [-11±√(289)]/42
a₁,₂ = [-11±17]/42
a₁ = 1/7
a₂ =-2/3
21a²+11a-2 = (7a-1)(3a+2)
e. 6a²+22a+20
Aplicar resolvente;
a₁,₂ = [-b±√(b²-4ac)]/2a
Sustituir;
a₁,₂ = [-22±√(22²-4(6)(20))]/2(6)
a₁,₂ = [-22±√(4)]/12
a₁,₂ = [-22±2]/12
a₁ = -5/3
a₂ =-2
6a²+22a+20 = 2(3a+5)(a+2)
f. 8a²-2a-15
Aplicar resolvente;
a₁,₂ = [-b±√(b²-4ac)]/2a
Sustituir;
a₁,₂ = [2±√(2²-4(30)(-21))]/2(8)
a₁,₂ = [2±√(484)]/16
a₁,₂ = [2±22]/16
a₁ = 3/2
a₂ =-5/4
8a²-2a-15 = (2a-3)(4a+5)
Respuesta:
Explicación paso a paso: