Answer:
[tex]y=x^4 + 4x^3 + 4x^2 + 4x + 3[/tex]
Step-by-step explanation:
Consider function [tex]y=x^4 + 4x^3 + 4x^2 + 4x + 3.[/tex]
Factor it
[tex]y=x^4 + 4x^3 + 4x^2 + 4x + 3\\ \\=x^4+x^3+3x^3+3x^2+x^2+x+3x+3\\ \\=x^3(x+1)+3x^2(x+1)+x(x+1)+3(x+1)\\ \\=(x+1)(x^3+3x^2+x+3)\\ \\=(x+1)(x^2(x+3)+(x+3))\\ \\=(x+1)(x+3)(x^2+1)[/tex]
This function has two real zeros [tex]x=-1[/tex] and [tex]x=-3[/tex] and two comples zeros (because [tex]x^2+1[/tex] cannot be factored futher).
Hence, the function [tex]y=x^4 + 4x^3 + 4x^2 + 4x + 3[/tex] has [tex]x=-1[/tex] and [tex]x=-3[/tex] as its only two real zeros.
