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Mathematics LibreTexts

3.3: Polynomial Equations

  • Page ID
    238
  • [ "article:topic", "authorname:green" ]

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    So far we have learned how to find the roots of a polynomial equation. If we have an equation that involves only polynomials we follow the steps:

    1. Bring all the terms over to the left hand side of the equation so that the right hand side of the equation is a 0.
    2. Get rid of denominators by multiplying by the least common denominator.
    3. If there is a common factor for all the terms, factor immediately.  Otherwise, multiply the terms out.
    4. Use a calculator to locate roots.
    5. Use the Rational Root Theorem and synthetic division to exactly determine the roots.

    Example 1

    Find all the rational solutions of 

    \[\dfrac{(2x^3 - 5)}{4} = x - x^2.\]

    Solution:

    1. \(\dfrac{(2x^3 - 5)}{4} - x + x^2 = 0\)

    2. \((2x^3 - 5)- 4x + 4x^2 = 0\)

    3. \(2x^3 + 4x^2 - 4x - 5 = 0\)

    4. From the graph, we see that there is a root between -3 and -2 and a root between 0 and -1 and a root between 1 and 2.  

    5. Since the only possible rational roots are 1, -1, 5, -5, .5, -.5, 2.5, -2.5, the possible rational roots are \(-\dfrac{5}{2}\) and  -.5.  Neither of these two are roots, hence there are no rational roots.

    Example 2

    Solve 

    \[x[x^2(2x + 3) + 10x + 17] + 5 = 2.\]

    1. \(x[x^2(2x + 3) + 10x + 17] + 3 = 0\)

    2. \(2x^4 + 3x^3 + 10x^2 + 17x + 3 = 0\)

    3. We see that there is a root between -2 and -1 and between -1 and 0.

    4. Our only possible roots are \(-\frac{1}{2}\) and \(-\frac{3}{2}\).

    5. Using synthetic division, we see that \(-\frac{3}{2}\) is a root, and the remainder is

      \[2x^3 + 10x + 2 = 2(x^3 + 5x + 1)\]

    which has no rational roots.  Hence the rational root is \(-\frac{3}{2}\) and using the calculator we see that the irrational root is 0.198.

    Larry Green (Lake Tahoe Community College)

    • Integrated by Justin Marshall.