Which of the following quartic functions has x=-1 and x = -2 as its only two real zeros?

Equation 3

Step-by-step explanation:

Lets see which of the functions has -2 as a zero root. We will go in order:

(1) (-2)^4 - 3(-2)^3 + 3(-2)^2 -3(-2) + 2 = 16 - 3(-8) + 3(4) + 6 +2 = 16 +24 +12 + 6 +2 =60 >0

So, (1) is wrong!

(2) (-2)^4 + 3(-2)^3 + 3(-2)^2 - 3(-2) - 2 = 16 - 24 + 12 + 6 - 2 =34 - 26 = 8 > 0

(2) is also wrong!

(3) (-2)^4 + 3(-2)^3 + 3(-2)^2 +3(-2) + 2 = 16 - 24 + 12 - 6 + 2 = 30 -30 = 0

The zero root x=-2 fits, what about x=-1?

(-1)^4 + 3(-1)^3 + 3(-1)^2 +3(-1) + 2 = 1 - 3 + 3 - 3 + 2 = 6 - 6 = 0

So, equation (3) fits both!

Finally, lets see (4):

(-2)^4 - 3(-2)^3 - 3(-2)^2 + 3(-2) + 2 = 16 + 24 - 12 - 6 + 2 = 42 - 18 = 24 > 0

So, (4) is also wrong.

Only equation 3 fits both zero roots!

since both sides equal, there are infinitely many olutions.

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !