91Ó°ÊÓ

Factoring quadratic expressions can seem tricky, but with practice, it becomes manageable. A quadratic expression is generally in the form of \(ax^2 + bx + c\). To factor it, you need to find two numbers that multiply to \(ac\) and add to \(b\). This step often involves trial and error along with some intuition.

Let's look at the quadratic expression from the problem: \(3x^2 + 8x - 3\).

• First, compute the product of \(a\) and \(c\), which is \(3 \times (-3) = -9\).
• Next, identify two numbers that multiply to \(-9\) and add to \(8\). These numbers are \(9\) and \(-1\).
• Rewrite the middle term using these two numbers: \(3x^2 + 9x - x - 3\).
• Group and factor by grouping: \[(3x^2 + 9x) + (-x - 3)\].
• Factor out common terms: \[3x(x + 3) - 1(x + 3)\].
• Now factor by using the common binomial factor: \[(3x - 1)(x + 3)\].

This breaking down into factors makes it much easier to work in other mathematical contexts like partial fraction decomposition.

When dealing with quadratic expressions, factoring them into linear factors is often the goal. Linear factors refer to polynomials of degree one, usually expressed as \(ax + b\). They are simpler to handle mathematically and provide an easier way to approach problems like partial fraction decomposition.

In the expression \(3x^2 + 8x - 3\), we faced the task of extracting linear factors. Once factored, it turned into \((3x - 1)(x + 3)\), each a linear factor.

• 3x - 1: This factor implies a simple line with a slope of 3 and a y-intercept of -1.
• x + 3: This too represents a line, with a slope of 1 and a y-intercept of 3.

Factoring into linear factors is crucial as it simplifies the expression, turning complex quadratic problems into more manageable linear ones. It allows further techniques such as partial fraction decomposition to be applied more straightforwardly.

Systems of equations come into play when solving for unknowns in multiple equations simultaneously. They often appear when breaking complex expressions into simpler parts, as seen in partial fraction decomposition.

In solving our problem, once we expressed the fraction with unknowns \(A\) and \(B\): \(-\frac{10}{(3x - 1)(x + 3)} = \frac{A}{3x - 1} + \frac{B}{x + 3}\), we needed to equate the numerators after clearing the denominators.

• The resulting equation \(-10 = A(x + 3) + B(3x - 1)\) expanded to \((A + 3B)x + (3A - B)\).
• By equating the coefficients of terms separately, we set up a system: \(A + 3B = 0\) and \(3A - B = -10\).
• Solving these simultaneously, we found \(A = -3\) and \(B = 1\).

The solutions \(A = -3\) and \(B = 1\) allow us to write the final partial-fraction decomposition. Mastery of systems of equations is fundamental, as it’s widely used in algebra to connect different parts of an equation to find unknown variables.