Final answer:
The quadratic expression 5y² - 10y - 5 can be factored by first taking out the common factor of 5, resulting in 5(y² - 2y - 1). Since the quadratic expression does not neatly factor with integers, we may use the quadratic formula if necessary, after which the expression may have irrational or complex roots.
Step-by-step explanation:
To factorize the quadratic expression 5y² - 10y - 5 using the second identity, we need to find two numbers that multiply to the product of the coefficient of y² (which is 5) and the constant term (which is -5), while also summing to the coefficient of y (which is -10). However, looking at the expression, we can first factor out the common factor of 5 from all terms, getting 5(y² - 2y - 1).
Now, we need to factor the quadratic expression in the parentheses. Since this expression does not factor neatly with integer values and it's not a difference of squares, we might use the quadratic formula to find its zeros as last resort:
- Rearrange the expression in the form of ax² + bx + c = 0, which is already done.
- Identify a = 1, b = -2, and c = -1.
- Apply the quadratic formula: y = (-b ± √(b² - 4ac))/(2a).
After applying the quadratic formula, we would get the two solutions for y, and we can then write our original expression as a product of two binomials, each one having one of the solutions as a root. However, those values are likely to be irrational or complex numbers, meaning the expression does not factor over the integers. In a typical high school setting, unless this is an introduction to complex numbers or irrational solutions, the most simplified factored form we'd provide would be 5(y² - 2y - 1), indicating that further factoring with real integers isn't possible.