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# State any two Euclid’s axioms and postulates. Also, find the remainder when $p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4$ is divided by $q\left( x \right)=1-2x$ .

Last updated date: 21st Jul 2024
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Hint: According to the question we are been asked to state any two axioms and postulates of Euclid’s and then we are been asked to find the remainder after dividing polynomial $p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4$by polynomial $q\left( x \right)=1-2x$.

Two Euclid’s axioms are:
(i) Things which are equal to the same are also equal with one another.
(ii) If equals are added to equals then holes are equal.
Postulates are:
(i) A circle can be drawn with any center and any radius.
(ii) A terminated line can be produced indefinitely.

Now, we are having polynomial $p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4$and we need to divide this with another polynomial which is $q\left( x \right)=1-2x$and hence attain the remainder as asked in the question.
Since, we know that the zero of q(x) is $\dfrac{1}{2}$ therefore, by using remainder theorem the remainder should be $f\left( \dfrac{1}{2} \right)$.
\begin{align} & \therefore f\left( \dfrac{1}{2} \right)={{\left( \dfrac{1}{2} \right)}^{3}}-6{{\left( \dfrac{1}{2} \right)}^{2}}+2\left( \dfrac{1}{2} \right)-4 \\ & \Rightarrow \dfrac{1}{8}-6\left( \dfrac{1}{4} \right)+1-4 \\ & \Rightarrow \dfrac{1}{8}-\dfrac{6}{4}-3 \\ \end{align}
Now taking the lcm and then solving this after simplifying we get $-\dfrac{35}{8}$ .
Therefore, the remainder after dividing p(x) from q(x) is $-\dfrac{35}{8}$.

Note: In this question we need some theory of Euclid’s and then we must know how to divide polynomials. We need to take care while dividing as we need to make the power the same and then cancel the terms and find the remainder which is although easy but may lead to some kind of careless mistakes.