Find all rational, irrational, and complex zeros (and state their multiplicities). $P ( x ) = x ^ { 4 } + x ^ { 3 } - 5 x ^ { 2 } - 3 x + 6$
A) $\{ - 3,3,1 , - 2 \}$
B) $\{ - 3,1 , - 2 \}$
C) $\{ - \sqrt { 3 } , \sqrt { 3 } , 1 , - \sqrt { 2 } \}$
D) $\{ - \sqrt { 3 } , \sqrt { 3 } , 1 , - 2 \}$
E) $\{ \sqrt { 3 } , - 6 \}$

Question

Quizplus · Accepted Answer

We can use either synthetic or long division to find that $(x - 1)$ is a factor of $P(x)$. Then, we have
$$P(x) = (x-1)(x^3 + 2x^2 - 3x -6)$$
Using the Rational Root Theorem, we can find that $x = -\sqrt{3}, \sqrt{3},$ and $-2$ are all roots of the cubic factor. Therefore, the roots of $P(x)$ are $- \sqrt{3}, \sqrt{3}, 1,$ and $-2.$ Since $P(x)$ is a fourth degree polynomial, it has $\boxed{	extbf{(D)} \{- \sqrt { 3 } , \sqrt { 3 } , 1 , - 2 \}}$ as its zeros.

Find All Rational, Irrational, and Complex Zeros (And State Their P(x)=x4+x3−5x2−3x+6P ( x ) = x ^ { 4 } + x ^ { 3 } - 5 x ^ { 2 } - 3 x + 6P(x)=x4+x3−5x2−3x+6

Find All Rational, Irrational, and Complex Zeros (And State Their $P ( x ) = x ^ { 4 } + x ^ { 3 } - 5 x ^ { 2 } - 3 x + 6$