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Upon receiving a question, it seemed to have needed me to use the remainder theorem however for a divisor that was not linear. Now while I could long divide it (or synthetically), I was wondering how $x-a$ would apply to something like $x^2-a$ or in general $x^k-a$. This would be a great help, thank you :)

N. F. Taussig
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John Hon
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  • Personally, I don't understand what you're trying to say. The description of the problem is fuzzy and imprecise. The only theorem about remainders that I can think of in which the "linearity" (degree = 1, I guess) of the divisor (?) comes into play is Ruffini's rule. Afterwards, when you say "$\text{stuff}$ applies to $\text{Stuff}$", how exactly would "$\text{stuff}$" apply to "$\text{Stuff}$" ? By "calculating the remainder of the polynomial divsion of $\text{Stuff}$ by $\text{stuff}$" ? [continues...] –  Aug 29 '16 at 09:40
  • [continued...] If so, how is this related to the issue in the previous statement (which looks like not being able to use Ruffini)? –  Aug 29 '16 at 09:40

2 Answers2

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To find the remainder, you substitute $x^{2} = a$ (resp. $x^{k} = a$) in the dividend, and collect the powers.

Andreas Caranti
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  • Could you apply it to $(x^3+5x-1):(x^2-1)?$ – mfl Aug 29 '16 at 09:45
  • Modulo $x^2 - 1$, one has $x^3 + 5 x - 1 = x \cdot x^2 + 5 x - 1 \equiv x + 5 x - 1 = 6 x - 1$. – Andreas Caranti Aug 29 '16 at 09:48
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    Understood. Thank you very much. +1 – mfl Aug 29 '16 at 09:53
  • I'm sorry I don't quite understand this solution and the one in the comments didnt exactly help either :(. If one was to substitute $x^2 = a$ then wouldn't you have $2a$? Im sorry, but could you use this as an example? Thanks a ton :)

    $(x^4+5x^3+2x+6)$ divided by $(x^2 -3)$

     – John Hon Aug 29 '16 at 13:07
  • I use "$\equiv$" to denote the substitution $x^{2} = 3$. Then we have $x^{4} + 5 x^{3} + 2 x + 6 = (x^2)^2 + 5 x \cdot x^2 + 2 x + 6 \equiv 9+ 15 x + 2 x + 6 = 17 x + 15$. – Andreas Caranti Aug 29 '16 at 13:12
  • Ah yes i see, however, is 17x+15 the final answer? and how does one get the substitution for x? is it just from setting $x^2 - 3 =0$ ? Thanks – John Hon Aug 29 '16 at 13:18
  • Yes, that's the remainder, you may check it with polynomial long division. I only substituted $3$ for $x^{2}$, no substitution for $x$ alone. – Andreas Caranti Aug 29 '16 at 13:44
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\begin{array}{c|c} \hline \text{Divisor} & \text{Remainder} \\ \hline x-a & f(a) \\[5pt] (x-a)(x-b) & \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b) \\[5pt] x^2-a^2 & \frac{f(a)-f(-a)}{2a}x+\frac{f(a)+f(-a)}{2} \\[5pt] (x-a)^{2} & (x-a)f'(a)+f(a) \\[5pt] (x-a)(x-b)(x-c) & \frac{(x-b)(x-c)}{(a-b)(a-c)} f(a)+ \frac{(x-c)(x-a)}{(b-c)(b-a)} f(b)+ \frac{(x-a)(x-b)}{(c-a)(c-a)} f(c) \\[5pt] (x-a)^{2} (x-b) & (x-a)^{2} \frac{f(b)-f(a)}{(b-a)^{2}}+ \frac{(x-a)(x-b)}{a-b} f'(a)+f(a) \\[5pt] (x-a)^{3} & \frac{(x-a)^{2}}{2}f''(a)+(x-a)f'(a)+f(a) \\ \hline \end{array}

Ng Chung Tak
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  • Is there any general case? – F Nishat Jan 29 '20 at 14:15
  • For divisor being a product of distinct linear factors, that's closely related to Lagrange interpolation. In case of power of linear factor, that's related to Taylor series expansion. Mixed cases are not trivial (or refer to Hermite interpolation for your further interest). – Ng Chung Tak Jan 29 '20 at 20:09