Use the grouping method to factor x3 + x2 + 3x + 3. A.(x + 1)(x2 + 3)
B.(x + 1)(x + 3)
C. x(x + 3)(x + 1)
D. (x + 1)(x+3)

(x+1) (x^2+3)

Step-by-step explanation:

x^3 + x^2 + 3x + 3

x^3 + x^2      + 3x + 3

Factor out x^2 from the first group and 3 from the second group

x^2(x+1)   +3(x+1)

Then factor out (x+1)

(x+1) (x^2+3)

Answer A is correct

Step-by-step explanation:

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yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value