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General forms of a polynomial \[ = a{x^2} + bx + c\]

A polynomial is denoted as \[p(x)\,or\,g(x)\]

A polynomial is always divided by another polynomial but with a lesser degree. By degree we mean the higher power of the variable in the respective polynomial i.e if \[p(x)\,\] is divided by some \[g(x)\] through long division then

Degree of \[g(x)\]

Given \[{x^3} + 3{x^2} + 3x + 1\] is divided by \[x\] i.e. let \[p(x) = {x^3} + 3{x^2} + 3x + 1.......(1)\] and \[g(x) = x\]

As we see that degree of \[g(x)\]i.e. I and degree of \[p(x)\] i.e. 3

\[ \Rightarrow \]degree of \[g(x) < \]degree of \[p(x)\]

Step1:- Make sure the polynomial is written is decreasing order. If any term is missing, use a zero to fill its place.

We have \[p(x) = {x^3} + 3{x^2} + 3x + 1\,and\,g(x) = x\]

Applying long division

\[x\sqrt {{x^3} + 3{x^2} + 3x + 1} ({x^2} + 3x + 3)\]

Step 2:- Dividing the \[g(x)\] with the highest power i.e \[x\] by \[{x^3}\] we get \[{x^2}\] we known \[x:{x^2} = {x^3}\]

Step 3:- Subtraitingit and bringing down the other terms.

Step 4: - Now to get \[3{x^2},x\] must be multiplied with \[3x\]. Hence by subtracting it we are left only with \[3x + 1\]

Step 5: - To get \[3x,x\] must be multiplied by 3, and by subtracting it we are left with 1 only.

Step 6: - We see that now, 1 is a constant and Hence out the remainder.

Therefore, we get 1 as a remainder on dividing \[{x^3} + 3{x^2} + 3x + 1\]by \[x\].

Remainder theorem:

If a polynomial \[p(x)\] is divided by \[x - a,\] the remainder is \[p(a)\]

Here as \[{x^3} + 3{x^2} + 3x + 1\]is divided by \[x\] or \[x + o\]

\[ \Rightarrow put\,x = 0\]in (1)

\[ \Rightarrow {(0)^3} + 3{(0)^2} + 3(0) + 1\]

\[\begin{gathered}

\Rightarrow 0 + 0 + 0 + 1 \\

\Rightarrow 1 \\

\end{gathered} \]

Hence the remainder is 1

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