Find all the zeros of f(x). f(x) = 2x3 + 7x2 - 28x + 12
Arrange your answers from smallest to largest. If
there is a double root, list it twice.

Plz help!

The zeroes are -6, 1/2 and 2.

Step-by-step explanation:

f(x) = 2x3 + 7x2 - 28x + 12 = 0

From the first and last coefficient 2 and 12, one guess for a zero is x = 2.

So substituting x = 2:

f(2) = 16+ 28 - 56 + 12 = 0

So x = 2 is a zero and x - 2 is a factor of f(x)

Performing long division:

x - 2)2x3 + 7x2 - 28x + 12( 2x2 + 11x - 6 < Quotient

       2x3  - 4x2

                 11x2 - 28x

                 11x2 - 22x

                          - 6x + 12

                           -6x + 12

Now we solve

2x2 + 11x - 6 = 0

(2x - 1)(x + 6) = 0

2x - 1 = 0 or x + 6 = 0, so:

x = 1/2,  x = -6.

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice