Question 1200922
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All the real zeros of the given polynomial are integers. Find the zeros. 
(Enter your answers as a comma-separated list. Enter all answers including repetitions.)
P(x) = x^3 − 3x − 2
x =
Write the polynomial in factored form.
P(x) =
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<pre>
It is an easy case of the Rational Zeroes theorem (since the leading coefficient 
of the polynomial is 1 (one)).


According to this theorem, the list of possible rational zeroes consists of these values
{1, -1, 2, -2}.


It is easy to check that x= -1 is the root.


Then the given polynomial is divisible by (x+1), so we divide the given polynomial
by (x+1) to reduce the degree

    {{{(x^3-3x-2)/(x+1)}}} = x^2 - x - 2.


Regarding the quadratic polynomial  x^2 - x - 2, we can factor it mentally

    x^2 - x - 2 = (x-2)*(x+1).


Therefore, the final decomposition of the given polynomial is 

    x^3 − 3x − 2 = {{{(x-2)*(x+1)^2}}}.


It has the roots x= 2 of multiplicity 1 and x= -1 of multiplicity 2.
</pre>

Solved.