Factor n^2+5n-24. Show that the original polynomial and the factored from describe the same sequence of numbers for n=0,1,2,3,and 4

Step-by-step explanation:

We have the original equation n ^ 2 + 5 * n - 24, when factoring we have:

(n + 8) * (n - 3)

Now by replacing the values:

n = 0

0 ^ 2 + 5 * 0 - 24 = - 24

(0 + 8) * (0 - 3) = - 24

n = 1

1 ^ 2 + 1 * 0 - 24 = - 18

(1 + 8) * (1 - 3) = -18

n = 2

2 ^ 2 + 5 * 2 - 24 = - 10

(2 + 8) * (2 - 3) = - 10

n = 3

3 ^ 2 + 5 * 3 - 24 = 0

(3 + 8) * (3 - 3) = 0

n = 4

4 ^ 2 + 5 * 4 - 24 = 12

(4 + 8) * (4 - 3) = 12

question:

could we have options?

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.