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I have a question related to writing a polynomial using powers of binomial of form $(x-a).$

I found an example: polynomial $P(x) = x^4 + 2x^3-3x^2-4x+1$ can be written as $ (x+1)^4-2(x+1)^3-3(x+1)^2+4(x+1)+1$ using powers of $(x+1)$ and Horner's Method. How do we obtain this representation of polynomial? How is Horner's Method used for that?

user121
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    The constant term in the expansion in powers of $x+1$ is the value at $x=-1$. Horner's method can be used to compute $P(-1)$. Moreover, in the calculation it also gives you the coefficients of $\frac{P(x)-P(-1)}{x+1}$ in the expansion in powers of $x$. So, you can compute the next coefficient in the expansion in powers of $x+1$ by evaluating this again at $x=-1$. Note how the $4$ in $4(x+1)$ is the constant term in the expansion of $\frac{P(x)-P(-1)}{x+1}$ in powers of $x+1$. – conditionalMethod Nov 03 '19 at 21:04

3 Answers3

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As you know you may get the result by the Taylor polynomial of $$ P(x) = x^4 + 2x^3-3x^2-4x+1$$ about $x=-1$

If you insist on the Horner's Method then it is be successive division by $x+1$

That is $$P(x) = P(-1) + P'(-1)(x+1) + P''(-1)(x+1)^2/2 +... $$

As you notice, the remainder in dividing your polynomial by $(x+1)$ is the constant which is $1$

Mohammad Riazi-Kermani
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You want an expression $a_4(x+1)^4+a_3(x+1)^3+a_2(x+1)^2+a_1(x+1)+a_0$. Repeatedly apply the method for $x=-1$. The last number in each row will give you the next coefficient starting from the lowest power. \begin{array}{|c|c|c|c|c|c|} \hline & 1 & 2 & -3 & -4 & 1 \\ \hline -1 & 1& 1& -4 & 0 & 1 \rightarrow a_0\\ \hline -1 & 1 & 0& -4 & 4 \rightarrow a_1\\ \hline -1 & 1 & -1& -3 \rightarrow a_2\\ \hline \end{array} and so on.

bjorn93
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This is similar to changing base of a number system.
So you can use long division. $$P(x) = x^4 + 2x^3-3x^2-4x+1$$
To express it as powers of $(x+1)$, divide $P(x)$ by the new base $(x+1)$

$$\begin{align} x^4 + 2x^3-3x^2-4x+1 &= (x+1)(x^3+x^2-4x) + \color{blue}{1} \\ x^3+x^2-4x &= (x+1)(x^2-4)+ \color{blue}{4}\\ x^2-4 &= (x+1)(x-1) \color{blue}{-3}\\ x-1 &= (x+1)(1) \color{blue}{- 2}\\ 1 &= (x+1)0+ \color{blue}{1}\\ \end{align}$$

$$P(x) = \color{blue}{1}(x+1)^4 \color{blue}{-2}(x+1)^3\color{blue}{-3}(x+1)^2+\color{blue}{4}(x+1)+\color{blue}{1}$$

AgentS
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