Question

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

Answer 1

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

Answer 2

$\frac{P(x)}{R(x)}|\frac{D(x)}{Q(x)}$

$\frac{x^4-2x^3+3x^2-10x+3}{}|\frac{x-2}{}$

$\frac{x^4-2x^3+3x^2-10x+3}{-x^4+2x^3}|\frac{x-2}{x^3}$

$\frac{3x^2-10x+3}{-3x^2+6x}|\frac{x-2}{x^3+3x}$

$\frac{-4x+3}{4x-8}|\frac{x-2}{x^3+3x-4}$

$\frac{-5}{}|\frac{x-2}{x^3+3x-4}$

$\boxed{\boxed{\frac{x^4-2x^3+3x^2-10x+3}{-5}|\frac{x-2}{x^3+3x-4}}}$

$R(x)=-5$