Question

Remainder Theorem

Answer 1

As requested by @ParthKohli,
\[\begin{align}
f(x)&=a_0+a_1x+a_2x^2+\dots+a_nx^n\\
g(x)&=b_0+b_1x+b_2x^2+\dots+b_mx^m\\
g(x)(x-a)&=(b_0+b_1x+b_2x^2+\dots+b_mx^m)(x-a)\\
&=b_0x+b_1x^2+b_2x^3+\dots +b_mx^{m+1}\\
&-(ab_0+ab_1x+ab_2x^2+\dots+ab_mx^m)\\
&=-ab_0+(b_0-ab_1)x+(b_1-ab_2)x^2\\
&+(b_2-ab_3)x^3+\dots+(b_{m-1}-ab_{m})x^{m}+b_mx^{m+1}\\
\text{Let } b_{i-1}-ab_{i}&=a_i \text{ where } i \in \{0,1,2,\dots,m\}\\
\text{Define } r(x)&=b_{-1}\\
\text{Then } g(x)(x-a)+r(x)&=(b_{-1}-ab_0)+(b_0-ab_1)x+(b_1-ab_2)x^2\\
&+(b_2-ab_3)x^3+\dots+(b_{m-1}-ab_{m})x^{m}+b_mx^{m+1}\\
\text{Since } b_{i-1}-ab_{i}&=a_i\\
\text{We can conclude }. . .\\
g(x)(x-a)+r(x)&=a_0+a_1x+a_2x^2+\dots+a_nx^n\\
&=f(x)\\
x=a &\Rightarrow r(a)=f(a).
\end{align}\]

Answer 2

hehe good luck with that @ParthKohli :)

Answer 3

I already understand it lol

Answer 4

The real part is after "We can conclude..."