To find the zeros of the function, we want to find the values of x when y is equal to zero.
So, we set the function equal to zero and solve for x:
x^3 - 5x^2 + 16x - 80 = 0
To solve this equation, we can use synthetic division or factor it by grouping.
Using synthetic division:
The constant term is -80, so the possible rational zeros are factors of 80, which are 1, 2, 4, 5, 8, 10, 16, 20, 40, and 80.
Let's start by checking 1:
1 | 1 -5 16 -80
| 1 -4 12
-----------------
1 -4 12 -68
Since the remainder is not zero, 1 is not a zero.
Now let's check -1:
-1 | 1 -5 16 -80
| -1 6 -22
-----------------
1 -6 22 -58
Again, the remainder is not zero, so -1 is not a zero.
Let's check 2:
2 | 1 -5 16 -80
| 2 -6 20
-----------------
1 -3 10 -60
Again, the remainder is not zero, so 2 is not a zero.
Let's check -2:
-2 | 1 -5 16 -80
| -2 14 -60
-----------------
1 -7 30 -40
Again, the remainder is not zero, so -2 is not a zero.
Let's check 4:
4 | 1 -5 16 -80
| 4 -4 48
-----------------
1 -1 12 -32
Again, the remainder is not zero, so 4 is not a zero.
Let's check -4:
-4 | 1 -5 16 -80
| -4 36 -80
-----------------
1 -9 52 0
Finally, we found a zero! -4 is a zero of the function.
Now that we have found a zero, we can use polynomial long division or synthetic division to divide the function by (x + 4):
(x + 4)(x^2 - 9x + 52)
Now we need to solve x^2 - 9x + 52 = 0.
This quadratic equation can be factored as:
(x - 4)(x - 13) = 0
Setting each factor equal to zero:
x - 4 = 0 or x - 13 = 0
Solving for x in each equation gives us:
x = 4 or x = 13
Therefore, the zeros of the function y = x^3 - 5x^2 + 16x - 80 are x = -4, x = 4, and x = 13.
Find all the zeros for the following function.
y=x^3-5x^2+16x-80
1 answer