Question

Diketahui suku banyak p(x) = x pangkat 4 + mx³ - 3x² + 2x +n . jika p(1)= 3 dan p(-1) = -5, tentukan :
a. nilai m dan n
b. nilai p (-2)

Answer 1

P(x) = x⁴ + mx³ - 3x² + 2x + n
p(1) = 3
1 + m - 3 + 2 + n = 3 
m + n = 3 ...(1)

p(-1) = -5
1 - m - 3 - 2 + n = -5
-m + n = -1 ...(2)

eliminasi (1) dan (2):
m + n = 3
-m + n = -1
_________ -
2m = 4
m = 2
n = 1

p(x) = x⁴ + 2x³ - 3x² + 2x + 1
p(-2) = (-2)⁴ + 2(-2)³ - 3(-2)² + 2(-2) + 1
= 16 - 16 - 12 - 4 + 1
= -15

Answer 2

P(x) = x^4 + mx³ - 3x² + 2x + n
 3   = 1^4 + m(1³) - 3(1²) + 2(1) + n
 3   = 1 + m - 3 + 2 + n  ⇔ m + n = 3 ...................................(1)

- 5  = (-1)^4 + m(-1)³ - 3(-1)² + 2(-1) + n
- 5  = 1 - m - 3 - 2 + n ⇔ -m + n = -1 ....................................(2)

Dari (1) dan (2)
 m + n = 3
-m + n = -1 +
       2n = 2 maka n = 1
m + n = 3 ⇒ m + 1 = 3 ⇔ m = 3-1 = 2
a) Jadi nilai m = 2 dan n = 1

P(x) = x^4 + 2x³ - 3x² + 2x + 1
b) p(-2) = (-2)^4 + 2(-2)³ -3(-2)² + 2(-2) + 1
           = 16 - 16 - 12 - 4 + 1
           = - 15