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The Rational Roots Test:
The zero of a polynomial is an input value (usually an x-value) that returns a value of zero for the
whole polynomial when you plug it into the polynomial. When a zero is a real (that is,
not complex) number, it is also an x-intercept of the graph of the polynomial function. You will
frequently (especially in calculus) want to know the location of the zeroes of a given polynomial
function. You could plug numbers into the polynomial, willy-nilly, and hope for the best. But as
you learned when you studied the Quadratic Formula, zeroes are often very messy numbers;
randomly guessing is probably not the best plan of attack. So how does one go about trying to
find zeroes?
The Rational Roots (or Rational Zeroes) Test is a handy way of obtaining a list of useful first
guesses when you are trying to find the zeroes (roots) of a polynomial. Given a polynomial with
integer (that is, positive and negative "whole-number") coefficients, the possible (or potential)
zeroes are found by listing the factors of the constant (last) term over the factors of the leading
coefficient, thus forming a list of fractions. This listing gives you a list of potential rational
(fractional) roots to test -- hence the name of the Test.
Let me emphasize: The Rational Roots Test does not give you the zeroes. It does not say what
the zeroes definitely will be. The Test only gives you a list of relatively easy and "nice" numbers
to try in the polynomial. Most of these possible zeroes will turn out not actually to be zeroes!
You can see the sense of the Test's methodology by looking at a simple polynomial. Given the
quadratic 12x2 – 7x – 10, you can use the Quadratic Formula to find the zeroes, but you can
also factor to get 12x2 – 7x – 10 = (3x + 2)(4x – 5). Setting the two factors equal to zero, you
get two roots at x = – 2/3 and x = 5/4. Note that the denominators "3" and "4" are factors of the
leading coefficiant "12", and the numerators "2" and "5" are factors of the constant term "10".
That is, the zeroes are fractions formed of factors of the constant term ( 10) over factors of the
leading coefficient (12). Note also, however, that fractions such as 5/6 and – 10/3 may also be
formed this way (and thus be provided to you by the Test), but these other fractions are not in fact
zeroes of this quadratic.
This relationship is always true: If a polynomial has rational roots, then those roots will be
fractions of the form (plus-or-minus) (factor of the constant term) / (factor of the leading
coefficient). However, not all fractions of this form are necessarily zeroes of the polynomial.
Indeed, it may happen that none of the fractions so formed is actually a zero of the polynomial.
Note that I keep saying "potential" roots, "possible" zeroes, "if there are any such roots...". This is
because the list of fractions generated by the Rational Roots Test is just a list
of potential solutions. It need not be true that any of the fractions is actually a solution. There
might not be any fractional roots! For example, given x2 – 2, the Rational Roots Tests gives the
following possible rational zeroes:
But you already know that:
...so the zeroes aren't actually rational at all.
Always remember: The Rational Roots Test only gives a list of good first guesses; it does NOT
give you "the" answers!

Find all possible rational x-intercepts of x4
+ 2x3 – 7x2 – 8x + 12.
The constant term is 12, with factors of 1, 2, 3, 4, 6, and 12. The leading coefficient in
this case is just 1, which makes my work a lot simpler. The Rational Roots Test says that
the possible zeroes are at:
Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved
= –12, –6, –4, –3, –2, –1, 1, 2, 3, 4, 6, 12
You can do a quick graph (especially if you have a
graphing calculator), and see that, out of the
above list, it would probably be good to start
looking for zeroes by plugging the values x = –3,
–2, 1, and 2into the polynomial.

Find all possible rational x-intercepts of y =
2x3 + 3x – 5.
Keeping in mind that x-intercepts are zeroes, I will use the Rational Roots Test.
The constant term of this polynomial is 5, with factors 1 and 5.
The leading coefficient is 2, with factors 1 and 2.
Then the Rational Roots Tests yields
the following possible solutions:
Don't forget the "plus-or-minus" on the solution. You either need to list out all the possible
solutions separately, as I did in the first example; or use a "plus-or-minus" in front of each
possible solution, as I showed here; or put one "plus-or-minus" in front of the whole list of
possible solutions. as I will show in the next example. Just make sure you have a "plus-or-minus"
in there somewhere.
By the way, as the graph shows, if there is a rational
root for y = 2x3 + 3x – 5, it has to be at x = 1.

Use the Rational Roots Test to find all possible rational zeroes of
6x4 – 11x3 + 8x2 – 33x – 30.
This problem will be more complicated than the previous one, because the leading
coefficient is not a simple "1". Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved
The constant term is 30, with factors 1,
2, 3, 5, 6, 10, 15, and 30.
The leading coefficient is 6, with factors 1,
Then the Rational Roots Test yields:
2, 3, and 6.
Check out the graph:
You can see from the graph that there may be rational roots at x = – 2/3 and x =
would probably not make sense to try any of the other listed potential zeroes.
+ 5/2, but it
Resource
Stapel, Elizabeth. "The Rational Roots Test: Introduction." Purplemath. Available from
http://www.purplemath.com/modules/rtnlroot.htm. Accessed 12 December 2010