yu mampir buat jawab soal matematika

[tex]\displaystyle f(x)=(x-2)(x+2)(x-3)h(x)+s(x)\\f(x)=(x-2)(x+2)(x-3)h(x)+ax^2+bx+c\\\\f(2)=(2-2)(2+2)(2-3)h(2)+a(2)^2+b(2)+c\\0=4a+2b+c\\\\f(-2)=(-2-2)(-2+2)(-2-3)h(-2)+a(-2)^2+b(-2)+c\\0=4a-2b+c\\\\f(3)=(3-2)(3+2)(3-3)h(3)+a(3)^2+b(3)+c\\0=9a+3b+c\\\\4a-2b+c=0\\4a+c=2b\\\\4a+2b+c=0\\2b+2b=0\\4b=0\\b=0\\\\4a+c=0\\c=-4a\\\\f(3)=9a+3b+c\\f(3)=9a-4a\\\frac15f(3)=a\\\\c=-4a\\c=-\frac45f(3)\\\\s(x)=ax^2+bx+c\\s(x)=\frac15f(3)x^2+0x-\frac45f(3)\\\boxed{\boxed{s(x)=\frac15f(3)(x^2-4)}}[/tex]

Mapel : Matematika
Kelas : XI SMA
Bab : Polinomial

Pembahasan :
f(x) suku banyak berderajat 5 habis dibagi (x² - 4) maka sisa 0

x² - 4
x = 2 ; x = -2

(x - 2)(x + 2)(x + 3)
x = 2 ; x = -2 ; x = -3

sisa pembagian f(x) oleh (x - 2)(x + 2)(x + 3) adalah s(x) = ax² + bx + c

S(x) = ax² + bx + c
S(x1) = s(-3) = 9a - 3b + c = 0
S(x2) = s(-2) = 4a - 2b + c = 0
S(x3) = s(2) = 4a + 2b + c = 0

Eliminasi...
4a + 2b + c = 0
4a - 2b + c = 0
————————— (-)
4b = 0
B = 0

S(x1) = 9a + c = f(3)
S(x2) = 4a + c = 0
S(x3) = 4a + c = 0

Eliminasi...
9a + c = f(3)
4a + c = 0
—————— (-)
5a = f(3)
a = 1/5 f(3)

Substitusi...
S(x1) = 9a + c = f(3)
S(x1) = 9(1/5 f(3)) + c = f(3)
9/5 f(3) + c = f(3)
C = f(3) - 9/5 f(3)
C = -4/5 f(3)

Maka...
S(x) = ax² + bx + c
S(x) = [1/5 f(3)] x² + [0]x + [4/5 f(3)]
S(x) = [1/5 f(3)] x² + [4/5 f(3)]
S(x) = 1/5 f(3) (x² - 4)