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How to Find the Roots of a Polynomial Equation Using Synthetic Division?

Published in Polynomial Roots 3 mins read

Synthetic division helps find roots of polynomial equations by simplifying the polynomial, enabling you to factor it and solve for the roots.

Steps to Find Roots Using Synthetic Division:

  1. Identify a Potential Root: Begin by identifying a potential root of the polynomial. You can often find this by using the Rational Root Theorem, which states that any rational root of the polynomial must be a factor of the constant term divided by a factor of the leading coefficient. Try testing these potential roots using synthetic division.

  2. Set Up Synthetic Division: Write the coefficients of the polynomial in a row. If any terms are missing (e.g., no x2 term), include a 0 as its coefficient. Place the potential root you are testing to the left.

  3. Perform the Division:

    • Bring down the first coefficient.
    • Multiply the potential root by the coefficient you just brought down.
    • Write the result under the next coefficient.
    • Add the two numbers together.
    • Repeat steps 3.2-3.4 until you reach the last coefficient.

  4. Check the Remainder: If the remainder (the last number in the bottom row) is 0, then the potential root you tested is indeed a root of the polynomial.

  5. Write the Resulting Polynomial: If the remainder is 0, the other numbers in the bottom row are the coefficients of the quotient polynomial, which has a degree one less than the original polynomial.

  6. Repeat if Necessary: If the quotient polynomial has a degree greater than 2, you can repeat the process to find additional roots. Keep going until you're left with a quadratic equation (degree 2) or a linear equation (degree 1).

  7. Solve for Remaining Roots: If you are left with a quadratic equation, you can use the quadratic formula or factoring to find the remaining roots. If you are left with a linear equation, you can solve it directly.

Example:

Let's say we want to find the roots of the polynomial: x3 + 6x2 + 11x + 6 = 0

  1. Potential Roots: Using the Rational Root Theorem, potential roots are ±1, ±2, ±3, ±6. Let's try -1.

  2. Synthetic Division Setup:

    -1 | 1   6   11   6
   |
   ------------------

  3. Perform Division:

    -1 | 1   6   11   6
   |    -1   -5   -6
   ------------------
     1   5    6   0

  4. Remainder Check: The remainder is 0, so -1 is a root.

  5. Resulting Polynomial: The quotient polynomial is x2 + 5x + 6 = 0

  6. Solve Remaining Roots: We can factor the quadratic: (x + 2)(x + 3) = 0. Thus, x = -2 and x = -3 are the other roots.

Therefore, the roots of x3 + 6x2 + 11x + 6 = 0 are -1, -2, and -3.

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