jika T(x) dibagi oleh (x^2-2x) dan (x^2+2x) berturut turut bersisa 4x-2 dan 3x+4 tentukan sisanya apabila T(x) dibagi (x^2-4)

[tex]t(x) = (x^{2} - 2x) \times q(x) + 4x - 2 \\ t(x) = x(x - 2) \times q(x) + 4x - 2 \\ t(2) = 2(2 - 2) \times q(2) + 4(2) - 2 = 0 + 8 - 2 = 6 \\ \\ t(x) = (x^{2} + 2x) \times q(x) + 3x + 4 \\ t(x) = x(x + 2) \times q(x) + 3x + 4 \\ t( - 2) = - 2( - 2 + 2) \times q( - 2) + 3( - 2) + 4 = 0 - 6 + 4 = - 2 \\ \\ t(x) = ( {x}^{2} - 4) + ax + b \\ t(x) = ( x - 2)(x + 2) + ax + b \\ t(2) = (2 - 2)(2 + 2) + 2a + b = 2a + b = 6 \\ t( - 2) = ( - 2 - 2)( - 2 + 2) - 2a + b = - 2a + b = - 2 \\ \\ 2a + b = 6 \\ b = 6 - 2a \\ - 2a + b = - 2 \\ - 2a + (6 - 2a) = - 4a + 6 = - 2 \\ - 4a = - 8 \\ a = 2 \\ b = 6 - 2a = 6 - 2(2) = 6 - 4 = 2 \\ \\ sisa = 2x + 2[/tex]
kalau ada yang kurang jelas bisa tanya