Math, asked by nanditanpatil14, 14 hours ago

In the North-East of India Flood affected a village. Mr. Sharma from Mumbai decided to help them with food items, clothes, medicines etc. so, he collected some amount from different persons, which is represented by 8x^4 + 14x^3 - 2x^2 + ax - b. If 4x^2 + 3x - 2 number of person were helped by this service and left over amount of 14x - 10 was donated in Prime Minister Relief Fund. Find the value of a and b. What values are possessed by Mr. Sharma?

Answers

Answered by gokulsanjayreddy
0

Answer:

2xsquare

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Answered by Swaramohite105
0

Answer:

a=1 and b=7

Step-by-step explanation:

Given,

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

4

+x

3

+8x

2

+ax+b)

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

2

+1)

As the total money collected given by (x^{2}+1)(x

2

+1) no of person.

so (x^{4}+x^{3}+8x^{2}+ax+b)(x

4

+x

3

+8x

2

+ax+b) is divisible by (x^{2}+1)(x

2

+1)

Let

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

4

+x

3

+8x

2

+ax+b)

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

2

+1)

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

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

Consider,

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

1

x

2

+b

1

x+c

1

So we can write the equation as

\begin{gathered}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}\end{gathered}

p(x)=g(x)×q(x)

⇒(x

4

+x

3

+8x

2

+ax+b)=(x

2

+1)(a

1

x

2

+b

1

x+c

1

)

⇒(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

)

⇒(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}x

4

on both side we get

a_{1}=1a

1

=1

By comparing co-efficient of x^{3}x

3

on both side we get

b_{1}=1b

1

=1

By comparing co-efficient of x^{2}x

2

on both side we get

\begin{gathered}a_{1}+c_{1}=8\\\\\Rightarrow{c_{1}}=8-a_{1}\\\\\Rightarrow{c_{1}}=8-1=7\end{gathered}

a

1

+c

1

=8

⇒c

1

=8−a

1

⇒c

1

=8−1=7

By comparing co-efficient of xx on both side we get

\begin{gathered}b_{1}=a\\\\\Rightarrow1=a\\\\\Rightarrow{a}=1\end{gathered}

b

1

=a

⇒1=a

⇒a=1

By comparing constant term of both side we get

\begin{gathered}b=c_{1}\\\\\Rightarrow{b}=7\end{gathered}

b=c

1

⇒b=7

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