Question

Three polynomials are factored below, but some coefficients and constants are missing. All of the missing values of a, b, c, and

d are integers.
1. x2 + 2x - 8 = (ax + b)(cx + d)
2. 2x3 + 2x? - 24x = 2x(ax + b)(cx + d)
3. 6x? - 15x - 9 = (ax + b)(ex + )

Answer 1

To factor the given polynomials, we use the quadratic formula to find the missing coefficients and constants.

Step-by-step explanation:

To factor the given polynomials, we can use the quadratic formula to find the missing coefficients and constants. Let's solve each equation:

1. x² + 2x - 8 = (ax + b)(cx + d)

   Using the quadratic formula, we have:

   x = (-b ± √(b² - 4ac)) / (2a)

   Comparing with the given polynomial, we can see that a = 1, b = 2, c = 4, and d = -2.

2. 2x³ + 2x? - 24x = 2x(ax + b)(cx + d)

   We need to find a, b, c, and d for this polynomial. By factoring out 2x from the equation, we get:

   2x(x² + 1 - 12) = 2x(ax + b)(cx + d)

   Comparing with the given polynomial, we can see that a = 1, b = 1, c = 1, and d = -6.

3. 6x?- 15x - 9 = (ax + b)(ex + )

   We need to find a, b, and e for this polynomial. By factoring out 3 from the equation, we get:

   3(x² - 5x - 3) = (ax + b)(ex + )

   Comparing with the given polynomial, we can see that a = 1, b = -5, and e = 3.