91Ó°ÊÓ

Factoring polynomials is a crucial step in partial fraction decomposition. It involves breaking down a polynomial into simpler, non-divisible factors that, when multiplied together, give the original polynomial. In our example, the polynomial given is: \(x^3 + x^2 - 2x\).
To factor this:

• First, identify the greatest common factor (GCF). Here, the GCF is \(x\). We factor it out: \(x(x^2 + x - 2)\).
• Next, factor the quadratic \(x^2 + x - 2\). This breaks down into two binomials: \((x + 2)(x - 1)\).
• Thus, the fully factored form of the polynomial is \(x(x + 2)(x - 1)\).

Understanding how to factor polynomials is key to setting up the partial fraction decomposition, as it simplifies the process of identifying the individual terms for which we're solving. Practice factoring various types of polynomials to master this foundational skill.

Linear factors are expressions of the form \(ax + b\), and they play a significant role in partial fraction decomposition. Given the denominator \(x(x + 2)(x - 1)\), we observe that it is composed of the linear factors \(x\), \(x + 2\), and \(x - 1\).
For each linear factor in the denominator, we can set up a fraction in our decomposition:

• \(\frac{A}{x}\)
• \(\frac{B}{x + 2}\)
• \(\frac{C}{x - 1}\)

These terms make it possible to combine the fractions and match numerators once we determine the constants \(A\), \(B\), and \(C\). Recognizing and working with linear factors is vital as it simplifies complex rational expressions and paves the way for easier algebraic manipulation and solution.

Solving algebraic equations is the final step to find the constants in partial fraction decomposition. Once we have set up our decomposed form:
\[\frac{4x^2 - 3x - 4}{x(x + 2)(x - 1)} = \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 1}\]
Multiply both sides by the denominator to eliminate the fractions. This gives us:
\[4x^2 - 3x - 4 = A(x + 2)(x - 1) + Bx(x - 1) + Cx(x + 2)\]
Expand and combine like terms from the right side to compare coefficients of corresponding powers of \(x\):
- For \(A\): Match terms involving \(x\) to solve \(A(x + 2)(x - 1)\).
- For \(B\): Match terms involving \(x\) to solve \(Bx(x - 1)\).
- For \(C\): Match terms involving \(x\) to solve \(Cx(x + 2)\).
To find the values, solve for \(A\), \(B\), and \(C\) by substitution (e.g., \(x = 0\), \(x = -2\), and \(x = 1\)) or by comparing coefficients. The solution often reveals:
\[A = -3, B = 1, C = 2\]
Finally, substitute these constants back into the partial fractions. This method sharpens your algebraic skills and helps in simplifying complex expressions into manageable parts.