Question

Find the remainder when \[{{x}^{3}}-a{{x}^{2}}+6x-a\] is divided by \[x-a\]?

Answer 1

Hint: We are given a polynomial expression, \[f(x)={{x}^{3}}-a{{x}^{2}}+6x-a\] and a monomial, \[g(x)=x-a\] and we are asked to divide the polynomial expression by the monomial so as to find the remainder of the given division operation. In order to find the remainder, we can use the remainder theorem. It states if \[f(x)\] a polynomial is divided by a monomial \[g(x)=x-a\], then the remainder is \[f(a)\]. So, we will substitute \[x=a\], in the given polynomial to find the value of\[f(a)\]. Hence, we will have the remainder of the given division operation.

Complete step-by-step solution:
According to the given question, we are given a polynomial equation and a monomial equation and we are asked to carry out its division so as to find the remainder of the same.
The given equations that we have is,
\[f(x)={{x}^{3}}-a{{x}^{2}}+6x-a\]
and \[g(x)=x-a\]
we have to carry out the division operation and find the remainder and for this we will make use of the Remainder Theorem. It states that \[f(x)\] a polynomial is divided by a monomial \[g(x)=x-a\], then the remainder is \[f(a)\].
In order to carry out the above mentioned, we will put \[x=a\] in the given polynomial expression and we will get the value of \[f(a)\], we get,
\[\Rightarrow f(a)={{a}^{3}}-a.{{a}^{2}}+6a-a\]
We get the new expression as,
\[\Rightarrow f(a)={{a}^{3}}-{{a}^{3}}+6a-a\]
We will now cancel out the terms with similar and appropriate signs and we have,
\[\Rightarrow f(a)=5a\]
Therefore, the remainder of the division of \[{{x}^{3}}-a{{x}^{2}}+6x-a\] by \[x-a\] is 5a.

Note: In the above solution, we made use of the remainder theorem, but we can also do the division operation using the long division method. But since, it wasn’t mentioned in the question to carry out the division operation using a specific method so we could go with any method we want. Also, while putting \[x=a\] in the polynomial equation, make sure no ‘x’ are left else, we will get ‘x’ in the remainder as well.