Answered

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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

Sagot :

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[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]
taijfreeman

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

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