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The Rational Roots Test: Introduction (page 1 of 2)

The Rational Roots (or Rational Zeroes) Test is a useful way to find your initial guesses when you are trying to find the zeroes (roots) of a polynomial. Given a polynomial with integer (positive and negative whole-number) coefficients, the possible (or potential) zeroes are formed by listing the factors of the constant (last) term over the factors of the leading coefficient, and forming a list of fractions. This gives you a list of potential rational (fractional) roots to test -- hence the name of the Test.

You can see the sense of this by looking at a simple quadratic. Given 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, but these other fractions are not zeroes of the original quadratic.

This is always true: If a polynomial has rational roots, then they 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:

    -2, -1, +1, +2

But you already know that:

    x = ± sqrt(2)

...so the zeroes aren't actually rational at all.

Always remember this: 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 our work a lot simpler. Then the Rational Roots Test says that the possible zeroes are at: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

      ± 1, 2, 3, 4, 6, 12

        = –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 by trying the values x = –3, –2, 1, and 2.

  

graph of y = x^4 + 2x^3 - 7x^2 - 8x + 12

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Cite this article as:

Stapel, Elizabeth. "The Rational Roots Test: Introduction." Purplemath. Available from
    http://www.purplemath.com/modules/rtnlroot.htm. Accessed
 

 

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