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**Hint:**For answering this question we will use the fact that the remainder when a polynomial $f\left( x \right)$ is divided by $\left( x-a \right)$ is given by $f\left( a \right)$ that is the point at which $\left( x-a \right)$ is equal to zero. And perform the further simplifications.

**Complete step by step answer:**

Now considering from the question we have the polynomial $4{{x}^{3}}-5{{x}^{2}}+6x-2$ which is divided by $x-1$ .

From the basic concept we know that the remainder when a polynomial is divided by another polynomial $x-a$ is given by setting the $x-a=0$ and then solve for $x$ and substituting the $x=a$ in the polynomial.

So here we need to substitute $x=1$ as the dividend is $x-1$ in the divisor $4{{x}^{3}}-5{{x}^{2}}+6x-2$.

After that we will have $4{{\left( 1 \right)}^{3}}-5{{\left( 1 \right)}^{2}}+6\left( 1 \right)-2$ .

After further simplifying we will have $4-5+6-2=10-7=3$ .

**Hence we can conclude that the remainder when $4{{x}^{3}}-5{{x}^{2}}+6x-2$ is divided by $x-1$ is $3$ .**

**Note:**While answering questions of this type we should take care with the calculations. For the case of polynomials division only the fact can be used that is when $f\left( x \right)$ is divided by $\left( x-a \right)$ the remainder will be given by $f\left( a \right)$ .

We have another method for solving that is the actual traditional division method. For this we need to divide $4{{x}^{3}}-5{{x}^{2}}+6x-2$ by $x-1$ .

By performing the division as follows, we will have

$\begin{align}

& \dfrac{4{{x}^{3}}-5{{x}^{2}}+6x-2}{x-1}=\dfrac{4{{x}^{2}}\left( x-1 \right)-{{x}^{2}}+6x-2}{x-1} \\

& \Rightarrow 4{{x}^{2}}+\dfrac{-{{x}^{2}}+6x-2}{x-1}=4{{x}^{2}}+\dfrac{-x\left( x-1 \right)+5x-2}{x-1} \\

& \Rightarrow 4{{x}^{2}}-x+\dfrac{5x-2}{x-1}=4{{x}^{2}}-x+\dfrac{5\left( x-1 \right)+3}{x-1} \\

& \Rightarrow 4{{x}^{2}}-x+5+\dfrac{3}{x-1} \\

\end{align}$

Hence the remainder when $4{{x}^{3}}-5{{x}^{2}}+6x-2$ is divided by $\left( x-1 \right)$ is $3$ . Hence in both the cases the answer is the same so any of the cases can be used but the first one is time efficient.

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