Synthetic Division (Ruffini’s Rule): worked exercises on polynomial division. A structured collection of step-by-step problems designed to master coefficient-based division, quotient extraction, and remainder analysis.
Each exercise follows a clear and systematic method: identification of the root, construction of the synthetic table, and interpretation of the final result.
Synthetic Division: Key Idea
Synthetic division is a simplified method for dividing a polynomial by a linear binomial of the form \(x - a\).
Instead of working with powers of \(x\), the method uses only the coefficients of the polynomial and performs a sequence of multiplications and additions.
The process is based on the identity:
\[ p(x) = (x - a)q(x) + r \]
where the remainder is given by:
\[ r = p(a) \]
Exercise 1
\[ (x^3 - 6x^2 + 11x - 6) : (x - 1) \]
Result
\[ x^2 - 5x + 6 \]
Solution
Identifying the root
\[ x - 1 = 0 \Rightarrow x = 1 \]
Coefficients
\[ 1 \quad -6 \quad 11 \quad -6 \]
Table
\[ \begin{array}{r|rrrr} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \end{array} \]
Conclusion
Quotient: \(x^2 - 5x + 6\), remainder \(0\).
Exercise 2
\[ (x^3 - 7x^2 + 14x - 8) : (x - 2) \]
Result
\[ x^2 - 5x + 4 \]
Solution
\(x=2\)
\[ \begin{array}{r|rrrr} 2 & 1 & -7 & 14 & -8 \\ & & 2 & -10 & 8 \\ \hline & 1 & -5 & 4 & 0 \end{array} \]
Exercise 3
\[ (x^3 + 2x^2 - 5x - 6) : (x + 3) \]
Result
\[ x^2 - x - 2 \]
Solution
\(x = -3\)
\[ \begin{array}{r|rrrr} -3 & 1 & 2 & -5 & -6 \\ & & -3 & 3 & 6 \\ \hline & 1 & -1 & -2 & 0 \end{array} \]
Exercise 4
\[ (x^3 - 2x^2 - 5x + 6) : (x + 2) \]
Result
\[ x^2 - 4x + 3 \]
Solution
\(x = -2\)
\[ \begin{array}{r|rrrr} -2 & 1 & -2 & -5 & 6 \\ & & -2 & 8 & -6 \\ \hline & 1 & -4 & 3 & 0 \end{array} \]
Exercise 5
\[ (2x^3 - x^2 - 7x + 6) : (x - 1) \]
Result
\[ 2x^2 + x - 6 \]
Solution
\[ \begin{array}{r|rrrr} 1 & 2 & -1 & -7 & 6 \\ & & 2 & 1 & -6 \\ \hline & 2 & 1 & -6 & 0 \end{array} \]
Exercise 6
\[ (x^3 - 9x^2 + 26x - 24) : (x - 3) \]
Result
\[ x^2 - 6x + 8 \]
Solution
\[ \begin{array}{r|rrrr} 3 & 1 & -9 & 26 & -24 \\ & & 3 & -18 & 24 \\ \hline & 1 & -6 & 8 & 0 \end{array} \]
Exercise 7
\[ (x^3 + 4x^2 - x - 4) : (x + 1) \]
Result
\[ x^2 + 3x - 4 \]
Solution
\(x=-1\)
\[ \begin{array}{r|rrrr} -1 & 1 & 4 & -1 & -4 \\ & & -1 & -3 & 4 \\ \hline & 1 & 3 & -4 & 0 \end{array} \]
Exercise 8
\[ (x^3 - 5x^2 - 2x + 24) : (x - 4) \]
Result
\[ x^2 - x - 6 \]
Solution
\[ \begin{array}{r|rrrr} 4 & 1 & -5 & -2 & 24 \\ & & 4 & -4 & -24 \\ \hline & 1 & -1 & -6 & 0 \end{array} \]
Exercise 9
\[ (x^3 + 3x^2 - 10x - 24) : (x + 4) \]
Result
\[ x^2 - x - 6 \]
Solution
\[ \begin{array}{r|rrrr} -4 & 1 & 3 & -10 & -24 \\ & & -4 & 4 & 24 \\ \hline & 1 & -1 & -6 & 0 \end{array} \]
Exercise 10
\[ (2x^3 + 3x^2 - 11x - 6) : (x - 2) \]
Result
\[ 2x^2 + 7x + 3 \]
Solution
\[ \begin{array}{r|rrrr} 2 & 2 & 3 & -11 & -6 \\ & & 4 & 14 & 6 \\ \hline & 2 & 7 & 3 & 0 \end{array} \]
Exercise 11–20
The same procedure applies systematically: identify the root \(a\), list the coefficients (including zeros for missing terms), construct the synthetic table, and read the quotient from the final row.
In all proposed exercises, the remainder is zero, confirming that the divisor is an exact factor of the polynomial.
Exercise 11
\[ (x^3 - x^2 - 14x + 24) : (x - 3) \]
Result
\[ x^2 + 2x - 8 \]
Solution
\(x=3\)
\[ \begin{array}{r|rrrr} 3 & 1 & -1 & -14 & 24 \\ & & 3 & 6 & -24 \\ \hline & 1 & 2 & -8 & 0 \end{array} \]
Exercise 12
\[ (x^3 - 4x^2 - 7x + 10) : (x + 2) \]
Result
\[ x^2 - 6x + 5 \]
Solution
\(x=-2\)
\[ \begin{array}{r|rrrr} -2 & 1 & -4 & -7 & 10 \\ & & -2 & 12 & -10 \\ \hline & 1 & -6 & 5 & 0 \end{array} \]
Exercise 13
\[ (3x^3 - 5x^2 - 4x + 4) : (x - 2) \]
Result
\[ 3x^2 + x - 2 \]
Solution
\(x=2\)
\[ \begin{array}{r|rrrr} 2 & 3 & -5 & -4 & 4 \\ & & 6 & 2 & -4 \\ \hline & 3 & 1 & -2 & 0 \end{array} \]
Exercise 14
\[ (x^3 + 5x^2 + 2x - 8) : (x + 4) \]
Result
\[ x^2 + x - 2 \]
Solution
\(x=-4\)
\[ \begin{array}{r|rrrr} -4 & 1 & 5 & 2 & -8 \\ & & -4 & -4 & 8 \\ \hline & 1 & 1 & -2 & 0 \end{array} \]
Exercise 15
\[ (2x^3 - 9x^2 + 7x + 6) : (x - 3) \]
Result
\[ 2x^2 - 3x - 2 \]
Solution
\(x=3\)
\[ \begin{array}{r|rrrr} 3 & 2 & -9 & 7 & 6 \\ & & 6 & -9 & -6 \\ \hline & 2 & -3 & -2 & 0 \end{array} \]
Exercise 16
\[ (x^3 - 7x + 6) : (x - 1) \]
Result
\[ x^2 + x - 6 \]
Solution
Coefficient nul à inclure : \[ 1 \quad 0 \quad -7 \quad 6 \]
\[ \begin{array}{r|rrrr} 1 & 1 & 0 & -7 & 6 \\ & & 1 & 1 & -6 \\ \hline & 1 & 1 & -6 & 0 \end{array} \]
Exercise 17
\[ (x^3 - 2x^2 - 9x + 18) : (x + 3) \]
Result
\[ x^2 - 5x + 6 \]
Solution
\(x=-3\)
\[ \begin{array}{r|rrrr} -3 & 1 & -2 & -9 & 18 \\ & & -3 & 15 & -18 \\ \hline & 1 & -5 & 6 & 0 \end{array} \]
Exercise 18
\[ (2x^3 + x^2 - 13x + 6) : (x + 3) \]
Result
\[ 2x^2 - 5x + 2 \]
Solution
\(x=-3\)
\[ \begin{array}{r|rrrr} -3 & 2 & 1 & -13 & 6 \\ & & -6 & 15 & -6 \\ \hline & 2 & -5 & 2 & 0 \end{array} \]
Exercise 19
\[ (x^3 + x^2 - 5x + 3) : (x - 1) \]
Result
\[ x^2 + 2x - 3 \]
Solution
\(x=1\)
\[ \begin{array}{r|rrrr} 1 & 1 & 1 & -5 & 3 \\ & & 1 & 2 & -3 \\ \hline & 1 & 2 & -3 & 0 \end{array} \]
Exercise 20
\[ (3x^3 + 2x^2 - 7x + 2) : (x - 1) \]
Result
\[ 3x^2 + 5x - 2 \]
Solution
\(x=1\)
\[ \begin{array}{r|rrrr} 1 & 3 & 2 & -7 & 2 \\ & & 3 & 5 & -2 \\ \hline & 3 & 5 & -2 & 0 \end{array} \]
Final Remark
In all exercises, the remainder is zero. Therefore, each divisor is an exact factor of the corresponding polynomial.
Synthetic division provides a fast and reliable method to compute the quotient and to identify polynomial factorizations.