Tolong bantu aku no 3 dan 4 beserta caranya

3)
p² – 4p – 5 = 0
(p – 5)(p + 1) = 0
p = 5 dan p = -1

koreksi soal
x³ – 2x² – 4x + 8 = 0
(x + 2)(x – 2)² = 0
x = -2 dan x = 2

x³ – 7x + 14x – 8 = 0
(x – 2)(x – 1)(x – 4) = 0
x = 2
x = 1
x = 4

x³ – 3x² – 4x = 0
x(x² – 3x – 4) = 0
x(x – 4)(x + 1) = 0
x = 0
x = -1
x = 4

4)
Qd = 3 – P²
Qs = 6P – 4

Qd = Qs
3 – P² = 6P – 4
P² + 6P – 4 – 3 = 0
P² + 6P – 7 = 0
(P + 7)(P – 1) = 0

P = -7 (tidak memenuhi)
P = 1

Q = 6P – 4
Q = 6(1) – 4
Q = 2

titik keseimbangan (2, 1)

Qd = 8 – P²
Qs = P² – 2

Qd = Qs
8 – P² = P² – 2
P² + P² = 8 + 2
2P² = 10
P² = 5
P² – 5 = 0
(P + √5)(P – √5) = 0

P = -√5 (tidak memenuhi)
P = √5 = 2,24

Q = P² – 2
Q = (√5)² – 2
Q = 5 – 2
Q = 3

titik keseimbangan (3, 2,24)