Find all the zeros of the function.

t^3 - 3t^2 - 15t + 125

Please explain. I am very confused on how to do this!!

Hint: In school problems, most are conjured to work in whole numbers. Notice the coefficents 3,5,125. That begs me to try 3, 5 as roots (experience).

So inspection I see that -5 is a root.
So now I would divide the polynomial by (t+5) either long, or synthetic. I get
t^2-8t+25 . You can find those factors with the quadratic formula...I don't think they will be real numbers. (b^2-4ac is negative)

Let

f(t)= t^3 - 3t^2 - 15t + 125
Then factorize f(t) as
f(t)=(t+5)*(t^2-8*t+25)
from which t=-5 is the real root.
The other factor of (t^2-8*t+25) yields two complex roots, since the discriminant is less than zero.
The two complex roots are obtained by solving:
(t^2-8*t+25) = 0
or t=4+3i or t=4-3i.

To find the zeros of the function, we need to find the values of t for which the function equals zero. One way to do this is by using the Rational Root Theorem.

Step 1: Determine the possible rational roots.
The Rational Root Theorem states that if a polynomial function has a rational root p/q, then p is a factor of the constant term (125 in this case), and q is a factor of the leading coefficient (1 in this case). Therefore, the possible rational roots are the divisors/factors of 125, which are ±1, ±5, ±25, ±125.

Step 2: Test the possible roots using synthetic division.
We can use synthetic division to test each possible rational root. We will start with the potential roots ±1.

For t = 1:
1 | 1 -3 -15 125
| 1 -2 -17
---------------
1 -2 -17 108

For t = -1:
-1 | 1 -3 -15 125
| -1 4 11
---------------
1 -4 -11 136

Neither ±1 produces a remainder of zero, so these are not zeros.

Step 3: Continue testing the remaining possible roots.
We will now test the remaining possible roots ±5, ±25, and ±125.

For t = 5:
5 | 1 -3 -15 125
| 5 10 -25
---------------
1 2 -5 100

For t = -5:
-5 | 1 -3 -15 125
| -5 40 -125
---------------
1 2 25 0

We found a remainder of zero when testing t = -5, which means -5 is a zero of the function.

Step 4: Write the factored form of the function.
Now that we know -5 is a root, we can use synthetic division again to divide the function by (t + 5).

-5 | 1 -3 -15 125
| -5 40 -125
---------------
1 2 25 0

This gives us a quotient of t^2 + 2t + 25. We can now write the original function as (t + 5)(t^2 + 2t + 25).

Step 5: Solve for the remaining zeros.
To find the zeros of t^2 + 2t + 25, we can use the quadratic formula or complete the square. However, in this case, the quadratic equation has no real solutions since the discriminant (b^2 - 4ac) is negative. Therefore, the remaining zeros are complex numbers and cannot be expressed as real values.

So, the zeros of the function t^3 - 3t^2 - 15t + 125 are -5, and two complex roots that cannot be expressed as real numbers.