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Find the remainder when \[{x^3} + 3{x^3} + 3x + 1\]is divided by \[x\].

Answer Verified Verified
Hint: Polynomials: - A polynomial is an expression consisting of variable and efficiency that involves only the operations of addition, subtraction, multiplication, etc.
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 \[g(x)\]Therefore,

Complete step by step answer:

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 \[x\].


Note: we can solve this question with the help of remainder theorem as well when degree of polynomial \[g(x) < \]degree of \[p(x)\]
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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