Latest TCS Aptitude Question SOLUTION: Find the number of polynomials of the form x^3+ax^2+bx+c that are divisible by x^2+1 ? where a,b,c belong to (1,2,3,4---------10).

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Find the number of polynomials of the form x^3+ax^2+bx+c that are divisible by x^2+1 ?
where a,b,c belong to (1,2,3,4---------10).

Asked In TCS pre (10 years ago)

Solved Rajarshi Bhattacharyya Read Solution (4)

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Rajarshi Bhattacharyya  (10 years ago)

A: x^3+ax^2+bx+c = (x^3+bx)+(ax^2+c).. now b has be 1 to get (x^2+1) as a common factor.. now a=1,c=1 is one solution.. hence its, (x^3+x)+(x^2+1)=x(x^2+1)+(x^2+1)=(x^2+1)(x+1) which is divisible by x^2+1..
values possible:
b=1, a=1, c=1
b=1, a=2, c=2
b=1, a=3, c=3
b=1, a=4, c=4
b=1, a=5, c=5
b=1, a=6, c=6
b=1, a=7, c=7
b=1, a=8, c=8
b=1, a=9, c=9
b=1, a=10, c=10
So.. Total of 10 solutions..

Is this solution Helpfull? Yes (73) | No (5)  

Gaurav Kaushik  (10 years ago)

10 solutions.
bcz when we divide by x^2+1 we get (b-1)x+(c-a) as remainder, and for x^2+1 to divide the given polynomial remainder should be zero. So to make remainder 0, b=1 and a=c, therefore 10 solutions possible.

Is this solution Helpfull? Yes (39) | No (11)  

Abhishek  (10 years ago)

10 solutions

Is this solution Helpfull? Yes (1) | No (19)  

Amit  (7 years ago)

on dividing the polynomial we get the remainder (b-1)x+(c-a) as it should be completely divisible so,
b-1=0 => b=1
and c-a=0 => c=a
so total 10 values will satisfy them as the range is given from 1,2,3...10.

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