Can someone please explain how and why Horner's method for dividing polynomials works? I was reading about it online but I don't even understand an example where $(x^3-6x^2+11x-6)$ is divided by $(x-2)$ let alone anything else.Thank you!
Explanation of Horner's method for dividing polynomials
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elementary-number-theory
polynomials
divisibility
1 Answers
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This method is based on the fact that if for a polynomial $P(x)$ we have that $P(a) = 0$ for some number $a$ then $P(x) = (x-a)Q(x)$ where the degree of $Q$ is one less than the degree of $P$. In your example $a=2$ since $P(2) = 0$ so $P(x) = (x-2)Q(x)$. Now $P(1) = 0$ so $(1-2)Q(1)=0$, so $Q(1) = 0$ giving $Q(x) = (x-1)R(x)$, but $P(3) = 0$ so $P(3) = (3-2) (3-1)R(3) = 0$. Since $R$ has degree $1$ $R(x) = (x-3)$. Conclusion $P(x) = (x-2)(x-1)(x-3)$.