Rational Zeros

Key Questions

  • Answer:

    See explanation...

    Explanation:

    The rational zeros theorem can be stated:

    Given a polynomial in a single variable with integer coefficients:

    #a_n x^n + a_(n-1) x^(n-1) + ... + a_0#

    with #a_n != 0# and #a_0 != 0#, any rational zeros of that polynomial are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #a_0# and #q# a divisor of the coefficient #a_n# of the leading term.

    Interestingly, this also holds if we replace "integers" with the element of any integral domain. For example it works with Gaussian integers - that is numbers of the form #a+bi# where #a, b in ZZ# and #i# is the imaginary unit.

  • The Rational Zeros Theorem states: If #P(x)# is a polynomial with integer coefficients and if #p/q# is a zero of #P(x)# , (# P(p/q) = 0 #), then #p# is a factor of the constant term of #P(x)# and q is a factor of the leading coefficient of P(x) .

  • Answer:

    See explanation...

    Explanation:

    A polynomial in a variable #x# is a sum of finitely many terms, each of which takes the form #a_kx^k# for some constant #a_k# and non-negative integer #k#.

    So some examples of typical polynomials might be:

    #x^2+3x-4#

    #3x^3-5/2x^2+7#

    A polynomial function is a function wholse values are defined by a polynomial. For example:

    #f(x) = x^2+3x-4#

    #g(x) = 3x^3-5/2x^2+7#

    A zero of a polynomial #f(x)# is a value of #x# such that #f(x) = 0#.

    For example, #x=-4# is a zero of #f(x) = x^2+3x-4#.

    A rational zero is a zero that is also a rational number, that is, it is expressible in the form #p/q# for some integers #p, q# with #q != 0#.

    For example:

    #h(x) = 2x^2+x-1#

    has two rational zeros, #x=1/2# and #x=-1#

    Note that any integer is a rational number since it can be expressed as a fraction with denominator #1#.

  • You can use the rational root theorem:

    Given a polynomial of the form:

    #a_0x^n+a_1x^(n-1)+...+a_n# with #a_0,...,a_n# integers,

    all rational roots of the form #p/q# written in lowest terms (i.e. with #p# and #q# having no common factor) will satisfy.

    #p | a_n# and #q | a_0#

    That is #p# is a divisor of the constant term and #q# is a divisor of the coefficient of the highest order term.

    This gives you a finite number of possible rational roots to try.

    For example, the rational roots of

    #6x^4-7x^3+x^2-7x-5=0#

    must be of the form #p/q# where #p# is #+-1# or #+-5# and
    #q# is #1#, #2#, #3# or #6#.

    You can try substituting each of the possible combinations of #p# and #q# as #x=p/q# into the polynomial to see if they work.

    In fact the only rational roots it has are #-1/2# and #5/3#.

    Once you have found one root, you can divide the polynomial by the corresponding factor to simplify the problem.

  • To find the zeroes of a function, #f(x)#, set #f(x)# to zero and solve.
    For polynomials, you will have to factor.

    For example: Find the zeroes of the function #f(x) = x^2+12x+32#

    First, because it's a polynomial, factor it
    #f(x) = (x+8)(x+4)#

    Then, set it equal to zero
    #0 = (x+8)(x+4)#

    Set each factor equal to zero and the answer is #x=-8# and #x=-4#.

    *Note that if the quadratic cannot be factored using the two numbers that add to this and multiple to be this method, then use the quadratic formula #(-b +- sqrt(b^2-4ac))/(2a)#

    to factor an equation in the form of #ax^2+bx+c#.

Questions