91Ó°ÊÓ

Polynomial factorization is a critical first step in breaking down expressions, like in our example with the expression \( \frac{2}{s^4 - 1} \). Start by recognizing the structure of the polynomial. Here, the term \( s^4 - 1 \) is a classic difference of squares. This type fits the form \( a^2 - b^2 \), which factors into \((a-b)(a+b)\). In this case, \( s^4 - 1 = (s^2)^2 - 1^2 \) becomes \( (s^2 - 1)(s^2 + 1) \).

Next, further factor \( s^2 - 1 \) since it is also a difference of squares: \((s - 1)(s + 1)\). The polynomial is now fully factored as \((s - 1)(s + 1)(s^2 + 1)\). Successfully breaking down the polynomial into its simpler components is key to applying partial fractions, enabling easier manipulation and resolution.

Remember:

• Identify the form of the polynomial (e.g., difference of squares).
• Factor it progressively until fully simplified.
• Use these factors in the following steps of partial fraction decomposition.

Once the polynomial is factored, it's time to set up and solve a system of equations. In partial fraction decomposition, each portion of the fraction corresponds to a piece of the factorized denominator. For our example: \[ \frac{2}{(s-1)(s+1)(s^2+1)} = \frac{A}{s-1} + \frac{B}{s+1} + \frac{Cs+D}{s^2+1} \].

To find variables \(A\), \(B\), \(C\), and \(D\), clear the fractions by multiplying through by the common denominator, giving:
\[ 2 = A(s+1)(s^2+1) + B(s-1)(s^2+1) + (Cs+D)(s^2-1) \].
Expand each term and collect like-power terms.

Setting up a system of equations from the coefficients of equivalent powers enables the resolution:

• Equate the coefficient of each power to the corresponding side of the equation ensuring equivalency.
• Formulate a system of simultaneous equations.
• Solve for unknowns \(A\), \(B\), \(C\), and \(D\).

This process allows us to determine specific values that simplify the original fraction expression.

After expanding and collecting terms, we use coefficient matching to find the values of unknowns in the system of equations. Coefficient matching involves equating the coefficients of like terms (from both sides of the equation) to solve for the unknowns.

For the decomposition:

• The equation can be expanded in terms of powers of \(s\):
  
  \[ 2 = (A + B + C)s^3 + (A - B + D)s^2 + (A + B - C)s + (A - B - D) \]}

From this, align coefficients:

• \(A + B + C = 0\)
• \(A - B + D = 0\)
• \(A + B - C = 0\)
• \(A - B - D = 2\)

Solving these equations step-by-step provides:

• From \(A + B + C = 0\) and \(A + B - C = 0\), \(2A = 0\) implies \(A = 0\).
• Substituting \(A = 0\) in the others, \(D = 1\)
• \(-B = 2\) so \(B = -2\)
• \(B = C\) gives \(C = -2\)

This systematic approach to finding coefficients is essential for reducing complex algebraic expressions to simpler, more interpretable forms.