Question

The expression \[ \frac{x^4}{(x^2+1)(x^2+3)} \] is equal to

• A. \(\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{x^2+3}\) for some \(A,B,C,D\in \mathbb{R}\setminus\{0\}\)
• B. \(\dfrac{Ax+B}{x^2+1}+\dfrac{Cx}{x^2+1}\) for some \(A,B,C\in \mathbb{R}\setminus\{0\}\)
• C. \(\dfrac{Ax}{x^2+1}+\dfrac{Bx}{x^2+3}\) for some \(A,B\in \mathbb{R}\setminus\{0\}\)
• D. \(1+\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{x^2+3}\) for some \(A,B,C,D\in \mathbb{R}\)

Hint:

If the degree of the numerator is equal to or greater than the degree of the denominator, first divide the polynomials. Then apply partial fractions to the proper rational part.

Answer 1

The Correct Option is D

Step 1: Observe the degrees of numerator and denominator.
The given expression is \[ \frac{x^4}{(x^2+1)(x^2+3)} \] First expand the denominator: \[ (x^2+1)(x^2+3)=x^4+4x^2+3 \] So the given expression becomes \[ \frac{x^4}{x^4+4x^2+3} \] Here, the degree of numerator is \[ 4 \] and the degree of denominator is also \[ 4 \]

Step 2: Since degrees are equal, divide first.
When the degree of numerator is greater than or equal to the degree of denominator, the rational expression is improper.
So, we first perform division: \[ x^4=(x^4+4x^2+3)-(4x^2+3) \] Hence, \[ \frac{x^4}{x^4+4x^2+3} = 1-\frac{4x^2+3}{x^4+4x^2+3} \] Since \[ x^4+4x^2+3=(x^2+1)(x^2+3), \] we get \[ \frac{x^4}{(x^2+1)(x^2+3)} = 1-\frac{4x^2+3}{(x^2+1)(x^2+3)} \]

Step 3: Apply partial fraction form.
Now the remaining fraction is \[ \frac{4x^2+3}{(x^2+1)(x^2+3)} \] Since both factors \[ x^2+1 \] and \[ x^2+3 \] are irreducible quadratic factors over \(\mathbb{R}\), the partial fraction form must be \[ \frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+3} \] Therefore, the whole expression must have the form \[ 1+\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+3} \] where \[ A,B,C,D\in \mathbb{R} \]

Step 4: Verify by solving constants.
Assume \[ \frac{x^4}{(x^2+1)(x^2+3)} = 1+\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+3} \] Multiplying throughout by \[ (x^2+1)(x^2+3), \] we get \[ x^4=(x^2+1)(x^2+3)+(Ax+B)(x^2+3)+(Cx+D)(x^2+1) \] Since \[ (x^2+1)(x^2+3)=x^4+4x^2+3, \] we have \[ 0=4x^2+3+(Ax+B)(x^2+3)+(Cx+D)(x^2+1) \] This can be satisfied for suitable real values of \(A,B,C,D\).
Thus, the correct form is option (4).

Step 5: Final conclusion.
Therefore, \[ \boxed{ 1+\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+3} } \]