Question

Find the remainder when \[{{x}^{3}}+{{x}^{2}}+x+1\] is divided by \[x-\dfrac{1}{2}\] by using the remainder theorem.

Answer 1

Hint: First of all try to recall the remainder theorem that is when f(x) is divided by x – c, then the remainder is f(c). Now, use this to find the required answer.

Complete step by step answer:
In this question, we have to find the remainder when \[{{x}^{3}}+{{x}^{2}}+x+1\] is divided by \[x-\dfrac{1}{2}\] 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,
\[f\left( x \right)=\left( x-c \right)q\left( x \right)+r\left( x \right)\]
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,
\[\Rightarrow f\left( x \right)=\left( x-c \right)q\left( x \right)+r\]
Now, when we substitute x = c, we get,
\[\Rightarrow f\left( c \right)=\left( c-c \right)q\left( c \right)+r\]
\[\Rightarrow f\left( c \right)=0+r\]
\[\Rightarrow f\left( c \right)=r\]
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 \[{{x}^{3}}+{{x}^{2}}+x+1\] so we get,
\[f\left( x \right)={{x}^{3}}+{{x}^{2}}+x+1......\left( i \right)\]
Here, we are dividing f(x) by \[x-\dfrac{1}{2}.\] So, we get \[c=\dfrac{1}{2}.\]
So, we get the remainder when f(x) is divided by x – c as
\[f\left( c \right)=f\left( \dfrac{1}{2} \right)\]
By substituting \[x=\dfrac{1}{2}\] in equation (i), we get,
\[\Rightarrow f\left( \dfrac{1}{2} \right)={{\left( \dfrac{1}{2} \right)}^{3}}+{{\left( \dfrac{1}{2} \right)}^{2}}+\left( \dfrac{1}{2} \right)+1\]
\[\Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{1}{8}+\dfrac{1}{4}+\dfrac{1}{2}+1\]
\[\Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{1+2+4+8}{8}\]
\[\Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{15}{8}\]

Hence, we get the remainder as \[\dfrac{15}{8}\] when \[{{x}^{3}}+{{x}^{2}}+x+1\] is divided by \[x-\dfrac{1}{2}.\]

Note: In this question, students must note that in x – c, they must take c with the sign like, for example, if we divide f(x) by x + 2, then we take c = – 2 since we can write x + 2 as x – (– 2). Also, we can verify our answer by dividing \[x-\dfrac{1}{2}\] by \[{{x}^{3}}+{{x}^{2}}+x+1\] using simple division and check if the remainder is \[\dfrac{15}{8}\] or not.