Question

# Total number of polynomials of the form $${ x }^{ 3 }+a{ x }^{ 2 }+bx+c$$ that are divisible by $${ x }^{ 2 }+1$$, where $$a,b,c\in \left\{ 1,2,3,......10 \right\}$$ is equal to

A
90
B
45
C
5
D
10

Solution

## The correct option is D $$10$$Let $$f\left( x \right) ={ x }^{ 3 }+a{ x }^{ 2 }+bx+c$$and $$g\left( x \right) ={ x }^{ 2 }+1$$Dividing $$f(x)$$ by $$g(x)$$$$\Rightarrow f\left( x \right) =g\left( x \right) (x+a)+(b-1)x+(c-a)$$Now if $$f(x)$$ is divisible by $$g(x)$$$$\Rightarrow (b-1)x+(c-a)=0$$$$\Rightarrow b-1=0$$  and $$c-a=0$$$$\Rightarrow b=1$$  and $$c=a$$As $$a$$ and $$c$$ belongs to $$\left\{ 1,2,3,4,5,6,7,8,9,10 \right\}$$So, $$a$$ and $$c$$ can be chosen in $$10$$ ways So, $$10$$ such polynomials are possible .So, option D is correct.Mathematics

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