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