Synthetic division is possible when the Divisior is in the form of $x+a$ or $x-a$. but what if the divisor is in the form of $x^2+a$, $x^2-a$, $x^3-a$,... and higher powers. how can we perform synthetic division in such cases.
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1Does this help? https://en.wikipedia.org/wiki/Synthetic_division#Expanded_synthetic_division – Jonas Meyer Feb 02 '17 at 02:09
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@@ΘΣΦ GenSan, what if the divisor is $x^2-a$ or $x^2+x+a$? – Aryabhatta Feb 02 '17 at 02:09
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##Jonas Meyer, what if the form is $x^2-a$? will it be $x=\pm \sqrt {a}$? – Aryabhatta Feb 02 '17 at 02:12
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I do not understand the question in the previous comment. You could divide separately by the linear factors though. The reference linked takes the polynomial as it is, even if not degree $1$. – Jonas Meyer Feb 02 '17 at 02:13
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@ Jonas Meyer, you mean once with $\sqrt {a}$ and again with $-\sqrt {a}$? – Aryabhatta Feb 02 '17 at 02:15
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@occasionaluser Are you interested in both quotient and remainder, or the remainder only? – dxiv Feb 02 '17 at 02:15
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@occasionaluser Yes, you could do that. – Jonas Meyer Feb 02 '17 at 02:16
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##dxiv, with both quotient as well as remainder . – Aryabhatta Feb 02 '17 at 02:18
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@occasionaluser Then you need to do either the long division, or the expanded synthetic division pointed out already. If you needed the remainder, only, then there are significant shortcuts. – dxiv Feb 02 '17 at 02:21
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@##dxiv, yeah I know that there is remainder theorem for finding the remainder only. but what is meant by expanded synthetic division. could you please show? – Aryabhatta Feb 02 '17 at 02:22
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@occasionaluser Follow the link posted by Jonas Meyer in the first comment. – dxiv Feb 02 '17 at 02:24
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@occasionaluser, you can do it by assuming $x^2=y$. – Feb 02 '17 at 02:26
1 Answers
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For the particular case of divisors in the form $x^n-a$ it is possible to replace the long division with $n$ synthetic divisions.
For a polynomial $P(x)$ being divided by $x^3-a$ for example, group the powers of $x$ in $P$ according to the remainder $\bmod 3$ and write it as:
$$ P(x) = P_0(x^3) + x P_1(x^3) + x^2P_2(x^3) $$
Use synthetic division to calculate the quotients and remainders of the following:
$$ P_k(x) = (x-a)Q_k(x) + r_k \quad\quad \text{for} \;\; k=0,1,2 $$
Then:
$$ P(x)=(x^3-a)Q(x) + R(x) $$
where $Q(x) = Q_0(x^3) + xQ_1(x^3)+x^2Q_2(x^3)$ and $R(x)=r_0+r_1x+r_2x^2\,$.
[ EDIT ] Following is a fully worked out example for $P(X)=x^4-6x^3+16x^2-25x+10$ (the polynomial was borrowed from another, unrelated question) being divided by $x^3-2$.
- Group the powers:
$$P(X)=x^4-6x^3+16x^2-25x+10= (-6x^3+10) + x\cdot (x^3-25) + x^2 \cdot 16$$
$$ \iff \begin{cases} \begin{align} P_0(x) & = -6x+10 \\ P_1(x) & = x - 25 \\ P_2(x) &= 16 \end{align} \end{cases} $$
- Divide $P_k$ by $x-2$ and determine $Q_k,r_k$ by synthetic division:
$$ \begin{cases} \begin{alignat}{3} P_0(x) & = -6x+10 && = -6(x-2) - 2\\ P_1(x) & = x - 25 && = (x-2) - 23\\ P_2(x) & = 16 && = 16 \end{alignat} \end{cases} $$
$$ \iff \begin{cases} \begin{align} Q_0(x) & = -6 \,,\;\; r_0 = -2\\ Q_1(x) & = 1 \,,\;\; r_1 = - 23\\ Q_2(x) & = 0 \,,\;\; r_2 = 16 \end{align} \end{cases} $$
- Calculate $Q,R$:
$$ Q(x) = Q_0(x^3) + xQ_1(x^3)+x^2Q_2(x^3) = -6 +x+ x^2 \cdot 0 = x-6\\ R(x)=r_0+r_1x+r_2x^2=16x^2-23 x-2\,$$
- Verify that indeed:
$$x^4-6x^3+16x^2-25x+10=(x^3-2)(x-6)+ 16x^2-23 x -2$$
dxiv
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