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Mathematics LibreTexts

3.3: Polynomial Equations

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.

Contributors

  • Integrated by Justin Marshall.