Use the REMAINDER THEOREM to explain whether or not (x-2) is a factor of F(x)=x^4-2x^3+3x^2-10x+3

[tex]\frac{P(x)}{R(x)}|\frac{D(x)}{Q(x)}[/tex][tex]\frac{x^4-2x^3+3x^2-10x+3}{}|\frac{x-2}{}[/tex][tex]\frac{x^4-2x^3+3x^2-10x+3}{-x^4+2x^3}|\frac{x-2}{x^3}[/tex][tex]\frac{3x^2-10x+3}{-3x^2+6x}|\frac{x-2}{x^3+3x}[/tex][tex]\frac{-4x+3}{4x-8}|\frac{x-2}{x^3+3x-4}[/tex][tex]\frac{-5}{}|\frac{x-2}{x^3+3x-4}[/tex][tex]\boxed{\boxed{\frac{x^4-2x^3+3x^2-10x+3}{-5}|\frac{x-2}{x^3+3x-4}}}[/tex][tex]R(x)=-5[/tex]

Answer:okStep-by-step explanation:\frac{P(x)}{R(x)}|\frac{D(x)}{Q(x)}\frac{x^4-2x^3+3x^2-10x+3}{}|\frac{x-2}{}\frac{x^4-2x^3+3x^2-10x+3}{-x^4+2x^3}|\frac{x-2}{x^3}\frac{3x^2-10x+3}{-3x^2+6x}|\frac{x-2}{x^3+3x}\frac{-4x+3}{4x-8}|\frac{x-2}{x^3+3x-4}\frac{-5}{}|\frac{x-2}{x^3+3x-4}\boxed{\boxed{\frac{x^4-2x^3+3x^2-10x+3}{-5}|\frac{x-2}{x^3+3x-4}}}R(x)=-5