This question would come in a unit on polynomial functions after the introduction of the Rational Root Theorem. The student would be expected to solve it this way:

A rational zero is a rational number, say p/q to lowest terms, that makes f(p/q) = 0.

The rational zero theorem says that p is a factor of the constant term of f, and q is a factor of the leading coefficient of f.

So all the possible rational zeros of f are:

factors of -20 ±1, ±2, ±4, ±5, ±10, ±20

--------------- => --------------------------------

factors of 1 ±1

=> ±1, ±2, ±4, ±5, ±10, and ±20 are the only possible rational zeros.

Now they need to be tested:

2 | 1 -4 1 16 -20

--- 2 -4 -6 20

------------------

1 -2 -3 10 | 0

----

So f(x) = (x - 2) (x^3 - 2 x^2 - 3 x + 10).

2 | 1 -2 -3 10

--- 2 0 -6

-------------

1 0 -3 | X => no double zero

----

-2 | 1 -2 -3 10

---- -2 8 -10

------------

1 -4 5 | 0

----

So f(x) = (x + 2) (x - 2) (x^2 - 4 x + 5).

For x^2 - 4 x + 5 = 0,

b^2 - 4ac = 16 - 20 = -4,

So f(x) has only two rational zeros: x = ±2

Notes: Rational implies Real. Real does not imply Rational. The fourth degree polynomial function