Which if the following quartic functions has x = 1 and x =

y = x⁴ +4x³ +4x² +4x +3

The coefficients of the offered quartics (in order) have 1, 1, 1, and 0 sign changes, respectively. Descartes' rule of signs tells you this means the first three choices all have one (1) positive real root, so the negative real roots -1 and -3 are not the only ones.

The only possible polynomial is the last one. Synthetic division of that polynomial by roots -1 and -3 leave the remaining factor as x²+1, which has only complex zeros.

The appropriate choice is ...

... y = x⁴ +4x³ +4x² +4x +3

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step-by-step explanation: