using the rational root theorem, state potential rational roots of P(x)=4x^3+x^2-7x+3?

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  • 2 months ago

    A rational number is a number that can be expressed as a "ratio" of integers.

    For example, 3/7 is a rational number.

    Often, we use the letters p and q to express the integers that will be used to write the rational number p/q

    (where q cannot be 0)

    The rational root theorem simply says that IF (a big if) the polynomial has rational roots, THEN they will be of the form

    p/q

    where p is an integer divisor of the constant term (the one without any x) and q is an integer divisor of the leading coefficient (the coefficient in front of the x with the highest degree)

    The constant term is 3.

    Therefore, p can be 1 or 3.

    The leading coefficient is 4.

    Therefore q can be 1, 2 or 4.

    The theorem says nothing about the sign (you have to try both versions: +(p/q) and -(p/q). There are other tricks to help you find the most likely sign, but the theorem cannot help you for the sign.

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    I usually pick from the shortest list and combine my choice with each one of the other list. For example, picking the "1" from the list for p, I can write:

    1/1 , 1/2 , 1/4

    (where 1/1 is simply "1")

    Then, using the 3, I can write

    3/1 , 3/2 , 3/4  (where 3/1 is simply "3")

    And you have to try each one as a positive value, then as a negative value.

    You then proceed by elimination.

    For example, we try x = 1

    4(1)^3 + (1)^2 - 7(1) + 3 = 4+1-7+3 = +1

    not zero, therefore 1 is not a root.

    Next, we try x = -1

    4(-1)^3 + (-1)^2 - 7(-1) + 3 = -4 + 1 + 7 + 3 = 7 (not zero, not a root)

    "But this could take time"

    Still a lot less time than trying random numbers out of an infinite list of ratios.

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    Because this is a "cubic" (a polynomial of degree 3), it cannot have more than 3 roots. If you find three values that work, then you can stop (you would have found all three).

    If you fail to find any, then it simply means that the polynomial does not have a rational root (it could still have roots that are of other kinds of numbers, like irrationals).

  • 2 months ago

    The integer factors of the constant term (3) are {1, 3}

    The integer factors of the leading coefficient (4) are {1, 2, 4}

    The rational root theorem states that potential rational roots will be of the from ±p/q, where p comes from the first set {1,3} and q comes from the second set {1, 2, 4}

    The first set has 2 elements, the second set has 3, so there should be 6 ways to combine them, or 12 if you count the ± sign.

    {±1/1, ±1/2, ±1/4, ±3/1, ±3/2, ±3/4} 

    Reduce them to lowest terms and remove any duplicates (there aren't any in this case, but if you had something like 4/8 and 1/2, you would get rid of one copy).

    {±1, ±1/2, ±1/4, ±3, ±3/2, ±3/4}

  • 2 months ago

    potential rational roots is a factor of the constant term over a factor of the high-degree coefficient.

    Going through the combinations, the list is:

    ±1, ±3, ±1/2, ±3/2, ±1/4, ±3/4

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