Newton-Raphson method for solving algebraic equations is based on

The correct option is Taylor Series.

The Newton-Raphson method is a powerful numerical technique used to find the roots of a real-valued function $f\left(x\right)=0$. The derivation of this method relies directly on the linear approximation of the function using its Taylor series expansion.

Let $x_n$ be an initial approximation to the root of $f\left(x\right)=0$, and let $h$ be a small correction such that the true root is at $x_{n+1}=x_n+h$. Therefore, we have:
$f(x_n+h)=0$

Expanding the function $f(x_n+h)$ about the point $x_n$ using the Taylor Series expansion gives:
$f(x_n+h)=f\left(x_n\right)+h{f}^{\prime }\left(x_n\right)+\frac{h^2}{2!}{f}^{″}\left(x_n\right)+...=0$

Assuming that $h$ is very small, we can neglect the higher-order terms involving $h^2$, $h^3$, and so on. This simplifies the expansion to a linear approximation:
$f\left(x_n\right)+h{f}^{\prime }\left(x_n\right)\approx 0$

Solving this linear equation for $h$, we obtain:
$h\approx -\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

Since $x_{n+1}=x_n+h$, we substitute the value of $h$ to find the next approximation:
$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

This is the standard iterative formula for the Newton-Raphson method, demonstrating that the technique is directly derived by truncating the Taylor Series expansion after the first-order derivative term.