Pembahasan :

f(x) = 2x⁴ - 8x² + 5x + 1

f(-4) = 2(-4)⁴ - 8(-4)² + 5(-4) + 1

= 365

f(4) = 2(4)⁴ - 8(4)² + 5(4) + 1

= 405

(3x² - 8x + 13)/((x +3)(x - 1)²) = A/(x+3) + B/(x-1) + C/(x-1)²

3x² - 8x + 13 = A(x-1)² + B(x+3)(x-1) + C(x+3)

3x²- 8x + 13 = Ax²-2Ax+A + Bx²+2Bx-3B + Cx+3C

3x² - 8x + 13 = (A+B)x²+(C -2A +2B)x +(A-3B+3C)

samakan kedua ruas:

A + B = 3

C - 2A + 2B = -8

A - 3B + 3C = 13

dengan eliminasi/substitusi dihasilkan :

A = 4, B = -1, C = 2

nilai A + B + C =

4 -1 +2 =

5

f(x) = x⁹ + ax⁶ + bx³ + a + b

habis dibagi (x²-1)

x² - 1 = 0

x² = 1

x = -1 atau x = 1

f(-1) = (-1)⁹ + a(-1)⁶ + b(-1)³ + a + b = 0

2a - 1 = 0

2a = 1

a = ½

f(1) = (1)⁹ + a(1)⁶ + b(1)³ + a + b = 0

2a+2b+1 =0

2(½)+2b+1=0

2b = -2

b = -1

jadi :

f(x) = x⁹ + ax⁶ + bx³ + a + b

f(x) = x⁹ + ½x⁶ + -x³ + ½ -1

f(x) = x⁹ +½x⁶- x³ - ½

berapa sisa jika dibagi (x³-x)

x³-x = x(x+1)(x-1) = 0

x = 0, x = 1, x = -1

f(x) = x⁹ +½x⁶- x³ -½

f(0) = -½

f(1) = 0

f(-1) = 0

sisa = ax²+bx+c

f(0) = a(0)²+b(0) +c

-½ = c

f(1) = a.1² + b(1) +c

0 = a + b + c

f(-1) = a(-1)²+b(-1)+c

0 = a -b +c

dengan eliminasi/substitusi didapat:

a = ½, b=0, c = -½

jadi sisa = ax² + bx + c

= ½x² + 0x -½

= ½x² - ½

Sisa = ½(x² - 1)