The zeros of the function p(x) = x2 – 2x– 24 are 1) -8 and 3
3) -4 and 6
2) -6 and 4
4) -3 and 8
Can u show work too

  3)  -4 and 6

Step-by-step explanation:

I find it easiest to solve these by factoring. Here, you're looking for factors of the constant (-24) that have a sum equal to the coefficient of the linear term (-2). You know the divisors of 24 are ...

  -24 = 1(-24) = 2(-12) = 3(-8) = 4(-6)

The sums of these factors are, respectively, -23, -10, -5, -2. So, the last pair of factors are the ones we're looking for. These are the constants that go into the binomial factors of the function:

  p(x) = x^2 -2x -24

  p(x) = (x +4)(x -6)

Then the zeros are the values of x that make these factors zero:

  x +4 = 0   ⇒   x = -4

  x -6 = 0   ⇒   x = 6

The zeros of the function are -4 and 6.

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The graph of the function confirms these values.

it would open into a square depending on the type of envelope i’m not sure about the second part