The equation x^3+x^2+ax-4=0 has one root equal to -2. find the value of a.

a = - 4

Step-by-step explanation:

Given x = - 2 is a root then f(- 2) = 0

f(x) = x³ + x² + ax - 4, thus

f(- 2) = (- 2)³ + (- 2)² - 2a - 4 = 0, that is

f(- 2) = - 8 + 4 - 2a - 4 = 0, thus

- 8 - 2a = 0 ( add 8 to both sides )

- 2a = 8 ( divide both sides by - 2 )

a = - 4

If n stands for the smallest number, then the second and third number would be (n + 1) and (n + 2)

Write the equation based on what are given
the sum of three numbers = 72
n + (n + 1) + (n + 2) = 72

Then, solve the equation
n + (n + 1) + (n + 2) = 72
3n + 3 = 72
3n = 72 - 3
3n = 69
n = 69/3
n = 23

The smallest number is 23