CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
45
loader
C
5
loader
D
10
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image