Can you find all real zeros of the function f(x) = x^3+5x^2-18x-72?

Answer 1

x^3+5x^2-18x-72

use trial and error with all factors p/q where p=-72 and q=1 (because q=1, this becomes more simple)

therefore, you find all factors of -72: 1, -1, 2, -2, 3, -3, 4, -4, and so on

you'll find that 4 is a root

4^3+5*4^2-18*4-72=64+80-72-72=144-144=0

so, x-4 is a factor

divide x-4 into the polynomial to get x^2+9x+18 which factors into (x+6)(x+3)

therefore x=-6 and x=-3 and we found that x=4 also

4, -6, and -3 are the real zeros

Answer 2

You can use synthetic division to find the first root.

1 5 -18 -72

-3 | 1 2 -24 0

So we can write x³ + 5x² - 18x - 72 = (x + 3)(x² + 2x -24)

We can easily factor (x² + 2x -24) as (x + 6)(x - 4)

So, the three real roots of f(x) are x = -6, x = -3, and x = 4