Answer:
"2"
Step-by-step explanation:
The quadratic equation has the form:
[tex]ay^2 + by + c[/tex]
Looking at the expression given, if we can replace "(x^3 + 1)" with another variable, such as u, then it will create a quadratic equation in the form shown above.
Let's replace "(x^3 + 1)" with u and check if it creates a quadratic:
[tex]16(x^3 + 1)^2 - 22(x^3 + 1) -3=0\\16u^2 -22u-3=0[/tex]
yes, this works because the equation we got in u falls in the quadratic category.
Answer choice 2 is right.