Answer :
Answer:
Option C
Step-by-step explanation:
Given expression is [tex]\frac{-5x^3+2x^2+9x+2}{(x^2+2)^2}[/tex]
Option A
[tex]\frac{A}{x^2+2}+ \frac{Bx+C}{(x^2+2)^2}[/tex]
= [tex]\frac{A(x^2+2)}{(x^2+2)^2}+ \frac{Bx+C}{(x^2+2)^2}[/tex]
= [tex]\frac{Ax^2+2A+Bx+C}{(x^2+2)^2}[/tex]
Numerator is a quadratic polynomial, while the expression has the numerator as a cubic .
Therefore, both the expressions are not equivalent.
Option B
[tex]\frac{A}{(x^2+2)^2}+ \frac{Bx+C}{(x^2+2)^2}[/tex]
= [tex]\frac{A+Bx+C}{(x^2+2)^2}[/tex]
Here, numerator of the expression is a linear polynomial which is not equal to given expression.
Therefore, both the expressions are not equivalent.
Option C
[tex]\frac{Ax+B}{(x^2+2)}+ \frac{Cx+D}{(x^2+2)^2}[/tex]
= [tex]\frac{(Ax+B)(x^2+2)}{(x^2+2)^2}+ \frac{Cx+D}{(x^2+2)^2}[/tex]
= [tex]\frac{Ax^3+Bx^2+2Ax+2B+Cx+D}{(x^2+2)^2}[/tex]
= [tex]\frac{Ax^3+Bx^2+x(2A+C)+D}{(x^2+2)^2}[/tex]
Here, numerator is a cubic polynomial.
Therefore, both the expressions are equivalent.
Option D
[tex]\frac{Ax+B}{(x^2+2)^2}+ \frac{Cx+D}{(x^2+2)^2}[/tex]
= [tex]\frac{x(A+C)+(B+D)}{(x^2+2)^2}[/tex]
Numerator is a linear polynomial.
Therefore, sum is not equal to the given expression.
Option C will be the answer.