[[1]] is the remainder when x3+x2+x+1 is divided by x-12  by using the remainder theorem.

# [[1]] is the remainder when x3+x2+x+1 is divided by $x-\frac{1}{2}$  by using the remainder theorem.

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### Solution:

In this question, we have to find the remainder when x3+x2+x+1
is divided by x−$\frac{1}{2}$
by using the remainder theorem. Before proceeding with this question, let us see what the remainder theorem is. Let us consider a polynomial f(x). When we divide f(x) by simple polynomial x – c, we get,
f(x)=(x−c)q(x)+r(x)
where q(x) is the quotient and r(x) is the remainder.
Here, the degree of (x – c) is 1. So, the degree of r(x) would be 0. So, we can say that r(x) is nothing but a constant r. So, we get,
⇒f(x)=(x−c)q(x)+r
Now, when we substitute x = c, we get,
⇒f(c)=(c−c)q(c)+r
⇒f(c)=0+r
⇒f(c)=r
So, from this, we can say that the remainder theorem states that when we divide a polynomial f(x) by x – c, the remainder is f(c).
Now, let us consider our question. Here, we are given a polynomial x3+x2+x+1
so we get,
f(x)=x3+x2+x+1......(i)
Here, we are dividing f(x) by x−$\frac{1}{2}$.
So, we get c=$\frac{1}{2}$.
So, we get the remainder when f(x) is divided by x – c as
f(c)=f($\frac{1}{2}$)
By substituting x = $\frac{1}{2}$  in equation (i), we get,
⇒f($\frac{1}{2}$)=()3+($\frac{1}{2}$)2+($\frac{1}{2}$)+1
⇒f($\frac{1}{2}$)= + $\frac{1}{4}$ + +1
⇒f($\frac{1}{2}$)=$\frac{1+2+4+8}{8}$
⇒f($\frac{1}{2}$)=$\frac{15}{8}$
Hence, we get the remainder as $\frac{15}{8}$ when x3+x2+x+1 is divided by x−.

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