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Factoring is a powerful technique used in solving quadratic equations. It involves expressing a quadratic equation in the form of a product of two binomials. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \). To factor it, you need to find two numbers that multiply to \( ac \) and add to \( b \). This can sometimes seem tricky, but with practice, it becomes more intuitive.

In the given exercise, the equation \( y^2 - 2y - 35 = 0 \) is already in standard form with \( a = 1 \), \( b = -2 \), and \( c = -35 \). Here, we look for two numbers that multiply to \(-35\) and add up to \(-2\). These numbers are \(5\) and \(-7\).

With these numbers, you can factor the quadratic equation as \((y - 7)(y + 5) = 0\). Each binomial can then be set to zero to find the values of \( y \). This process reveals the potential solutions for the quadratic equation.

The substitution method is a strategy to simplify complex equations by changing variables. This technique is particularly useful when dealing with awkward exponents or complex expressions.

In our exercise, we substitute \( y = x^{-1} \), which simplifies the original problem. The difficult negative exponent \(x^{-2}\) transforms into \(y^2\), creating a standard quadratic equation \(y^2 - 2y - 35 = 0\). By making the substitution, we simplify the process significantly, allowing us to focus on solving the quadratic form.

Once the quadratic equation is solved for \( y \), we then substitute back to find the original values of \( x \). This step-by-step simplification helps in managing complex algebraic problems.

Verification is an essential step in any math problem-solving process. This stage ensures that the solutions obtained satisfy the original equation.

For the exercise at hand, after determining values for \( y \) and converting them to \( x \) using the relationship \( y = x^{-1} \), solutions \( x = \frac{1}{7} \) and \( x = -\frac{1}{5} \) are verified. Verification involves substituting these solutions into the original equation \( x^{-2} - 2 x^{-1} - 35 = 0 \).

Plugging \( x = \frac{1}{7} \) back, and doing the same for \( x = -\frac{1}{5} \), confirms whether each value holds true, keeping the equation balanced. This ensures accuracy and confirms that the steps taken were correct, a vital part of any solution.