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!

sas (side angle side)

aa(angle angle)

sss(side side side)

sas you have to find two sides with an angle in between

aa you find two angles because it will always add up to 180

sss find all sides

step-by-step explanation: