Question

The sum of coefficients of the polynomial is equal to 2 and the sum of coefficients in even places is equal to the sum of coefficients in odd places.
Find the remainder of dividing f(x) by x^2-1.

Answer 1

Refer this attachment

Hope it's help you

Answer 2

Concept

This problem is based on binomial expansion which is used to expand and write the terms which are equal to the natural number exponent of the sum or difference of two terms. Let two numbers are $x$ and $y$ then the binomial expansion to the power of n is

$(x+y)^n={}^{n}C_{0}x^ny^0+{}^{n-1}C_{1}x^{n-1}y^1+{}^{n}C_{2}x^{n-2}y^2+...+{}^{n}C_{n}x^0y^{n}$

Given

We have given that the sum of coefficients of the polynomial is equal to 2 and the sum of coefficients in even places is equal to the sum of coefficients in odd places.

To Find

We have to find the remainder of dividing  $f(x)$  by $x^2-1.$

Solution

The binomial expansion:$(x+y)^n={}^{n}C_{0}x^ny^0+{}^{n-1}C_{1}x^{n-1}y^1+{}^{n}C_{2}x^{n-2}y^2+...+{}^{n}C_{n}x^0y^{n}$

Let $f(x)$ be polynomials

Put $y=1$ , then the expansion will be

$(x+1)^n={}^{n}C_{0}x^n+{}^{n-1}C_{1}x^{n-1}+{}^{n}C_{2}x^{n-2}+...+{}^{n}C_{n}x^0$

We have given the sum of coefficients of the polynomial is equal to 2

i.e. ${}^{n}C_{0}+{}^{n-1}C_{1}+{}^{n}C_{2}+...+{}^{n}C_{n}=2$

But from properties of binomial coefficients when we put $x=1$

${}^{n}C_{0}+{}^{n-1}C_{1}+{}^{n}C_{2}+...+{}^{n}C_{n}=2^n$

and the sum of coefficients in even places is equal to the sum of coefficients in odd places

i.e. $C_{0}+C_{2}+C_{4}+...=C_{1}+C_{3}+C_{5}+...$

But from properties of binomial coefficients

$C_{0}+C_{2}+C_{4}+...=C_{1}+C_{3}+C_{5}+...=2^{n-1}$

Since polynomial $f(x)$ is divided by  $x^2-1$ i.e  $(x-1)(x+1)$

When we divide a polynomial $f(x)$   by $x-a$ then the remainder is $f(a)$

Here, $a=1$  and also the sum of coefficients of the polynomial is equal to $2$

So, then the remainder is   $f(1)=2$

When we divide a polynomial $f(x)$   by $x+a$  then the remainder is $f(-a)$

So, then the remainder is   $f(-1)=0$

Polynomial $f(x)$ can be written like this,

$f(x)=Q(x)(x^2-1)+kx+b$

Where   $Q(x)$  is the quotient and $kx+b$ is the remainder

Also  $f(1)=2=k+b$

and  $f(-1)=0=-k+b$

So, we get $k=0$  and $b=2$  i.e. the remainder is $2$

As a result, the remainder of dividing   $f(x)$  by $x^2-1$ is $2$.

#SPJ3