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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

User Nourah
by
7.6k points

1 Answer

4 votes

Answer:

3) -4 and 6

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.

____

The graph of the function confirms these values.

The zeros of the function p(x) = x2 – 2x– 24 are 1) -8 and 3 3) -4 and 6 2) -6 and-example-1
User Stefano Dalpiaz
by
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