Question

$( {x}^{3} - 8 {x}^{2} + 19x - 9) \div (x - 4)$

Answer 1

$(x^3-8x^2+19x-9)\div(x-4)=x^2-4x+3\\\underline{-x^3+4x^2}\\=-4x^2+19x\\\underline{\ \ \ \ \ 4x^2-16x}\\.\ \ \ =3x-9\\\underline{\ \ \ \ \ -3x+12}\\.\ \ \ \ \ =\ 3$

$x^3-8x^2+19x-9=(x^2-4x+3)(x-4)+3$