Question

A village in the north east in India is suffering from flood. A merchant decided to help them with food items etc. So he collected some money from different persons which is represented by x4+x3+8x2+ax+b. If the number of persons who contributed for this service are x2+1 find the value of a and b

Answer 1

Answer:

a=1 and b=7

Step-by-step explanation:

Given,

Total amount of money collected = $(x^{4}+x^{3}+8x^{2}+ax+b)$

No of person = $(x^{2}+1)$

As the total money collected given by $(x^{2}+1)$ no of person.

so $(x^{4}+x^{3}+8x^{2}+ax+b)$ is divisible by $(x^{2}+1)$

Let

$p(x)=(x^{4}+x^{3}+8x^{2}+ax+b)$

$g(x)=(x^{2}+1)$

Since $g(x)$ divides $p(x)$

therefore the quotient $q(x)$ is a polynomial of degree 2

Consider,

$q(x)=a_{1}x^{2}+b_{1}x+c_{1}$

So we can write the equation as

$p(x)=g(x)\times{q}(x)\\\\\Rightarrow(x^{4}+x^{3}+8x^{2}+ax+b)=(x^{2}+1)(a_{1}x^{2}+b_{1}x+c_{1})\\\\\Rightarrow(x^{4}+x^{3}+8x^{2}+ax+b)=(a_{1}x^{4}+b_{1}x^{3}+c_{1}x^{2}+a_{1}x^{2}+b_{1}x+c_{1})\\\\\Rightarrow(x^{4}+x^{3}+8x^{2}+ax+b)=a_{1}x^{4}+b_{1}x^{3}+(c_{1}+a_{1})x^{2}+b_{1}x+c_{1}$

By comparing co-efficient of $x^{4}$ on both side we get

$a_{1}=1$

By comparing co-efficient of $x^{3}$ on both side we get

$b_{1}=1$

By comparing co-efficient of $x^{2}$ on both side we get

$a_{1}+c_{1}=8\\\\\Rightarrow{c_{1}}=8-a_{1}\\\\\Rightarrow{c_{1}}=8-1=7$

By comparing co-efficient of $x$ on both side we get

$b_{1}=a\\\\\Rightarrow1=a\\\\\Rightarrow{a}=1$

By comparing constant term of both side we get

$b=c_{1}\\\\\Rightarrow{b}=7$