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Understanding how to factor quadratic expressions is a vital skill in algebra. A quadratic expression is generally in the form of \(ax^2 + bx + c\). Factoring involves finding two binomials that multiply to give the original quadratic expression. In the given exercise, we need to factor the expression \(2x^2 + x - 15\) to solve the equation by finding a common denominator.

To factor \(2x^2 + x - 15\), we look for two numbers that multiply to \(a \times c = 2 \times (-15) = -30\) and add up to \(b = 1\). These numbers are \(6\) and \(-5\). Using these numbers, we decompose the middle term: \(2x^2 + 6x - 5x - 15\). Grouping terms gives \((2x^2 + 6x) + (-5x - 15)\).

Factor each group: \(2x(x + 3) - 5(x + 3)\). We can now see the common factor \((x + 3)\), leading to \((2x - 5)(x + 3)\). This factoring is crucial as it provides a common denominator for the fractions involved.

In rational equations, like the one in this exercise, finding a common denominator is crucial to combine fractions effectively. It's similar to finding a common denominator when adding fractions in arithmetic.

For the equation \[\frac{11}{2x^2 + x - 15} = \frac{5}{2x-5} - \frac{x}{x+3},\] we determine the least common denominator by factoring the quadratic expression as \((2x-5)(x+3)\). This makes the solution process more manageable, as all fractions are converted to have this common denominator.

Once rewritten with a common denominator, we can effectively combine the fractions on the right side of the equation because they share the same base \((2x - 5)(x + 3)\). This eliminates the need for the fractions during the solving process.

The quadratic formula is a mathematical tool that provides a solution for any quadratic equation of the form \(ax^2 + bx + c = 0\). This universal formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]

This formula is particularly useful when factoring is difficult or when a quadratic does not easily decompose into integer factors. In the exercise, after equating numerators, we transformed the equation into \(x^2 - 5x - 2 = 0\).

Here, \(a = 1\), \(b = -5\), and \(c = -2\). By substituting these into the quadratic formula, we find the solutions are \(x = \frac{5 \pm \sqrt{33}}{2}\). This means there are two possible solutions for \(x\): one using the plus sign and the other using the minus sign.

The discriminant is a vital part of the quadratic formula, denoted as \(b^2 - 4ac\). It determines the type and nature of roots of the quadratic equation \(ax^2 + bx + c = 0\).

In the context of the exercise, after manipulating the equation to \(x^2 - 5x - 2 = 0\), we calculate the discriminant as \((-5)^2 - 4 \times 1 \times (-2) = 25 + 8 = 33\).

If the discriminant is:

• Positive, there are two distinct real roots.
• Zero, there is one repeated real root.
• Negative, there are two complex roots.

For our equation, since the discriminant is \(33\) (a positive number), we discover two distinct real solutions. This result is critical for understanding the behavior of the quadratic graph and solutions.