Answer
$x^4-2x^3-5x+4$
Work Step by Step
As we are given that $f(x)=x^4-2x^3-5x+4$
Consider the Maclaurin Series of order $n$ for the function $f(x)$ at the point $a=0$ as follows:
$f(x)=f(0)+\Sigma_{n=1}^\infty \dfrac{f^{n}(0)}{n!}(x-0)^n+..$
Now, we have $f(0)=0-0-0+4=4 \\ f'(x)=4x^3-6x^2-5 \implies f'(0)=0-0-5=-5 \\ f''(x)=12x^2-12x \implies f''(0)=12(0)-12(0)=0\\ f'''(x)=24x-12 \implies f'''(0)=24(0)-12 =-12\\f^{4}(x)=24 \implies f^{4}(0)=24 \\f^{n}(x)=0 \implies f^{n}(0)=0$
Therefore, $f(x)=f(0)+\Sigma_{n=1}^\infty \dfrac{f^{n}(0)}{n!}(x-0)^n=f(0)+\Sigma_{n=1}^\infty \dfrac{f^{n}(0)}{n!}(x-0)^n=4-5x-\dfrac{12}{3!}x^3+\dfrac{24}{4!}x^4$
Thus, $f(x)=x^4-2x^3-5x+4$
