Answer:
option D
Step-by-step explanation:
x=-1 and x=-3 are the real zeros
We plug in the x values in each equation and check which equation makes y=0
A) y=x^4-4x^3-4x^2-4x-3
x=-1 , [tex]y=(-1)^4-4(-1)^3-4(-1)^2-4(-1)-3= 2[/tex]
x=-3 , [tex]y=(-3)^4-4(-3)^3-4(-3)^2-4(-3)-3= 162[/tex]
y is not equal to 0 so option A is not correct
B.y=-x^4-4x^3+4x^2+4x+3
x=-1 , [tex]y=(-1)^4-4(-1)^3+4(-1)^2+4(-1)+3=6 [/tex]
x=-3 , [tex]y=(-3)^4-4(-3)^3+4(-3)^2+4(-3)+3=54 [/tex]
y is not equal to 0 so option B is not correct
C.y=x^4+4x^3+3x^2+4x-4
x=-1 , [tex]y=(-1)^4+4(-1)^3+4(-1)^2+4(-1)-4= -8[/tex]
x=-3 , [tex]y=(-3)^4+4(-3)^3+4(-3)^2+4(-3)-4= -16[/tex]
y is not equal to 0 so option C is not correct
D) y=x^4+4x^3+4x^2+4x+3
x=-1 , [tex]y=(-1)^4+4(-1)^3+4(-1)^2+4(-1)+3= 0[/tex]
x=-3 , [tex]y=(-3)^4+4(-3)^3+4(-3)^2+4(-3)+3= 0[/tex]
y=0 so option D has two real zeros x=-1 and x=-3
