Learn to divide polynomials using Ruffini's rule. Complete exercises with tables, detailed steps, and fully worked solutions for high school students.
Exercise 1:
\[ (x^3 - 6x^2 + 11x - 6) : (x - 1) \]
Result
\[ x^2 - 5x + 6 \]
Solution
Approach
The divisor is a linear binomial of the form \(x - a\): in such cases Ruffini's rule is the most efficient tool available. Rather than carrying out polynomial long division in full, one works exclusively with the coefficients of the dividend, significantly reducing the number of operations required.
Finding the key value
The divisor is \(x - 1\). Ruffini's rule makes use of the value that makes the divisor equal to zero: \[ x - 1 = 0 \;\Rightarrow\; x = 1 \] Therefore: \[ a = 1 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 - 6x^2 + 11x - 6 \] The corresponding coefficients are: \[ 1 \quad -6 \quad 11 \quad -6 \]
Setting up the synthetic division table
Place the value \(a = 1\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} 1 & 1 & -6 & 11 & -6 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = 1\) and place the result beneath the next coefficient:
\[ 1 \cdot 1 = 1 \]
Step 3: add the values in the column:
\[ -6 + 1 = -5 \]
Step 4: repeat the process: multiply the value obtained by \(1\) and add:
\[ -5 \cdot 1 = -5 \] \[ 11 + (-5) = 6 \]
Step 5: carry out the final step:
\[ 6 \cdot 1 = 6 \] \[ -6 + 6 = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 - 5x + 6 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 - 5x + 6} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 1\) and can be written as: \[ x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) \]
Exercise 2:
\[ (x^3 - 7x^2 + 14x - 8) : (x - 2) \]
Result
\[ x^2 - 5x + 4 \]
Solution
Approach
We recognise that the divisor is a linear binomial of the form \(x - a\). This allows us to apply Ruffini's rule: rather than performing polynomial long division column by column, it suffices to arrange the coefficients of the dividend in a table and carry out an alternating sequence of multiplications and additions.
Finding the key value
The divisor is \(x - 2\), already in standard form \(x - a\): no rewriting is necessary. The value that makes the divisor zero follows directly: \[ x - 2 = 0 \;\Rightarrow\; x = 2 \] Therefore: \[ a = 2 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 - 7x^2 + 14x - 8 \] The corresponding coefficients are: \[ 1 \quad -7 \quad 14 \quad -8 \]
Setting up the synthetic division table
Place the value \(a = 2\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} 2 & 1 & -7 & 14 & -8 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = 2\) and place the result beneath the next coefficient:
\[ 1 \cdot 2 = 2 \]
Step 3: add the values in the column:
\[ -7 + 2 = -5 \]
Step 4: repeat the process: multiply the value obtained by \(2\) and add:
\[ -5 \cdot 2 = -10 \] \[ 14 + (-10) = 4 \]
Step 5: carry out the final step:
\[ 4 \cdot 2 = 8 \] \[ -8 + 8 = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 2 & 1 & -7 & 14 & -8 \\ & & 2 & -10 & 8 \\ \hline & 1 & -5 & 4 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 - 5x + 4 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 - 5x + 4} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 2\) and can be written as: \[ x^3 - 7x^2 + 14x - 8 = (x - 2)(x^2 - 5x + 4) \]
Exercise 3:
\[ (x^3 + 2x^2 - 5x - 6) : (x + 3) \]
Result
\[ x^2 - x - 2 \]
Solution
Approach
The divisor is a linear binomial. Before applying Ruffini's rule, careful attention must be paid to the sign: the divisor is not of the form \(x - a\) with \(a\) positive, but rather \(x + 3\). It is therefore essential to correctly identify the root of the divisor — that is, the value of \(x\) that makes it zero — so as to avoid sign errors throughout the process.
Finding the key value
The divisor is \(x + 3\). Since Ruffini's rule requires the form \(x - a\), we rewrite the divisor making the sign explicit: \[ x + 3 = x - (-3) \] The value that makes the divisor zero is therefore: \[ x + 3 = 0 \;\Rightarrow\; x = -3 \] Therefore: \[ a = -3 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 + 2x^2 - 5x - 6 \] The corresponding coefficients are: \[ 1 \quad 2 \quad -5 \quad -6 \]
Setting up the synthetic division table
Place the value \(a = -3\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} -3 & 1 & 2 & -5 & -6 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = -3\) and place the result beneath the next coefficient:
\[ 1 \cdot (-3) = -3 \]
Step 3: add the values in the column:
\[ 2 + (-3) = -1 \]
Step 4: repeat the process: multiply the value obtained by \(-3\) and add:
\[ -1 \cdot (-3) = 3 \] \[ -5 + 3 = -2 \]
Step 5: carry out the final step:
\[ -2 \cdot (-3) = 6 \] \[ -6 + 6 = 0 \]
Complete table
\[ \begin{array}{r|rrrr} -3 & 1 & 2 & -5 & -6 \\ & & -3 & 3 & 6 \\ \hline & 1 & -1 & -2 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 - x - 2 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 - x - 2} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x + 3\) and can be written as: \[ x^3 + 2x^2 - 5x - 6 = (x + 3)(x^2 - x - 2) \]
Exercise 4:
\[ (x^3 - 2x^2 - 5x + 6) : (x + 2) \]
Result
\[ x^2 - 4x + 3 \]
Solution
Approach
Here again the divisor is linear with a positive constant term, exactly as in the previous exercise. Recall that Ruffini's rule requires inserting into the table not the constant term of the divisor, but its root — the value of \(x\) that makes it zero. A sign error at this stage would propagate through every subsequent step.
Finding the key value
The divisor is \(x + 2\). Rewriting in the form \(x - a\): \[ x + 2 = x - (-2) \] The value that makes the divisor zero is: \[ x + 2 = 0 \;\Rightarrow\; x = -2 \] Therefore: \[ a = -2 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 - 2x^2 - 5x + 6 \] The corresponding coefficients are: \[ 1 \quad -2 \quad -5 \quad 6 \]
Setting up the synthetic division table
Place the value \(a = -2\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} -2 & 1 & -2 & -5 & 6 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = -2\) and place the result beneath the next coefficient:
\[ 1 \cdot (-2) = -2 \]
Step 3: add the values in the column:
\[ -2 + (-2) = -4 \]
Step 4: repeat the process: multiply the value obtained by \(-2\) and add:
\[ -4 \cdot (-2) = 8 \] \[ -5 + 8 = 3 \]
Step 5: carry out the final step:
\[ 3 \cdot (-2) = -6 \] \[ 6 + (-6) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} -2 & 1 & -2 & -5 & 6 \\ & & -2 & 8 & -6 \\ \hline & 1 & -4 & 3 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 - 4x + 3 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 - 4x + 3} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x + 2\) and can be written as: \[ x^3 - 2x^2 - 5x + 6 = (x + 2)(x^2 - 4x + 3) \]
Exercise 5:
\[ (2x^3 - x^2 - 7x + 6) : (x - 1) \]
Result
\[ 2x^2 + x - 6 \]
Solution
Approach
The divisor is of the form \(x - a\), so Ruffini's rule applies directly. It is worth noting that the leading coefficient of the dividend is \(2\), not \(1\): this presents no obstacle whatsoever, since the rule operates on the coefficients exactly as they are, regardless of their value. All that is needed is to enter them correctly in the table.
Finding the key value
The divisor is \(x - 1\), already in standard form \(x - a\). The value that makes the divisor zero is: \[ x - 1 = 0 \;\Rightarrow\; x = 1 \] Therefore: \[ a = 1 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ 2x^3 - x^2 - 7x + 6 \] The corresponding coefficients are: \[ 2 \quad -1 \quad -7 \quad 6 \]
Setting up the synthetic division table
Place the value \(a = 1\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} 1 & 2 & -1 & -7 & 6 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 2 \]
Step 2: multiply the value just written by \(a = 1\) and place the result beneath the next coefficient:
\[ 2 \cdot 1 = 2 \]
Step 3: add the values in the column:
\[ -1 + 2 = 1 \]
Step 4: repeat the process: multiply the value obtained by \(1\) and add:
\[ 1 \cdot 1 = 1 \] \[ -7 + 1 = -6 \]
Step 5: carry out the final step:
\[ -6 \cdot 1 = -6 \] \[ 6 + (-6) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 1 & 2 & -1 & -7 & 6 \\ & & 2 & 1 & -6 \\ \hline & 2 & 1 & -6 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial. Since the dividend was degree 3 with leading coefficient \(2\), the quotient is degree 2 with leading coefficient \(2\): \[ 2x^2 + x - 6 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{2x^2 + x - 6} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 1\) and can be written as: \[ 2x^3 - x^2 - 7x + 6 = (x - 1)(2x^2 + x - 6) \]
Exercise 6:
\[ (x^3 - 9x^2 + 26x - 24) : (x - 3) \]
Result
\[ x^2 - 6x + 8 \]
Solution
Approach
The divisor is of the form \(x - a\), so we apply Ruffini's rule. It is useful to recall that the degree of the quotient is always exactly one less than the degree of the dividend: when dividing a degree-3 polynomial by a degree-1 binomial, we expect a degree-2 quotient. This allows us to sanity-check the result at a glance.
Finding the key value
The divisor is \(x - 3\), already in standard form \(x - a\). The value that makes the divisor zero is: \[ x - 3 = 0 \;\Rightarrow\; x = 3 \] Therefore: \[ a = 3 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 - 9x^2 + 26x - 24 \] The corresponding coefficients are: \[ 1 \quad -9 \quad 26 \quad -24 \]
Setting up the synthetic division table
Place the value \(a = 3\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} 3 & 1 & -9 & 26 & -24 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = 3\) and place the result beneath the next coefficient:
\[ 1 \cdot 3 = 3 \]
Step 3: add the values in the column:
\[ -9 + 3 = -6 \]
Step 4: repeat the process: multiply the value obtained by \(3\) and add:
\[ -6 \cdot 3 = -18 \] \[ 26 + (-18) = 8 \]
Step 5: carry out the final step:
\[ 8 \cdot 3 = 24 \] \[ -24 + 24 = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 3 & 1 & -9 & 26 & -24 \\ & & 3 & -18 & 24 \\ \hline & 1 & -6 & 8 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 - 6x + 8 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 - 6x + 8} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 3\) and can be written as: \[ x^3 - 9x^2 + 26x - 24 = (x - 3)(x^2 - 6x + 8) \]
Exercise 7:
\[ (x^3 + 4x^2 - x - 4) : (x + 1) \]
Result
\[ x^2 + 3x - 4 \]
Solution
Approach
The divisor is \(x + 1\), a special case in which the root is \(-1\). In Ruffini's rule, this means that at every step one multiplies by \(-1\) — that is, one simply changes the sign of the current value before adding it to the next coefficient. This is computationally convenient, but still requires care: the alternating signs can lead to errors if one works too hastily.
Finding the key value
The divisor is \(x + 1 = x - (-1)\). The value that makes the divisor zero is: \[ x + 1 = 0 \;\Rightarrow\; x = -1 \] Therefore: \[ a = -1 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 + 4x^2 - x - 4 \] The corresponding coefficients are: \[ 1 \quad 4 \quad -1 \quad -4 \]
Setting up the synthetic division table
Place the value \(a = -1\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} -1 & 1 & 4 & -1 & -4 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = -1\) and place the result beneath the next coefficient:
\[ 1 \cdot (-1) = -1 \]
Step 3: add the values in the column:
\[ 4 + (-1) = 3 \]
Step 4: repeat the process: multiply the value obtained by \(-1\) and add:
\[ 3 \cdot (-1) = -3 \] \[ -1 + (-3) = -4 \]
Step 5: carry out the final step:
\[ -4 \cdot (-1) = 4 \] \[ -4 + 4 = 0 \]
Complete table
\[ \begin{array}{r|rrrr} -1 & 1 & 4 & -1 & -4 \\ & & -1 & 3 & -4 \end{array} \]
\[ \begin{array}{r|rrrr} -1 & 1 & 4 & -1 & -4 \\ & & -1 & -3 & 4 \\ \hline & 1 & 3 & -4 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 + 3x - 4 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 + 3x - 4} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x + 1\) and can be written as: \[ x^3 + 4x^2 - x - 4 = (x + 1)(x^2 + 3x - 4) \]
Exercise 8:
\[ (x^3 - 5x^2 - 2x + 24) : (x - 4) \]
Result
\[ x^2 - x - 6 \]
Solution
Approach
The presence of a linear binomial as the divisor calls for Ruffini's rule. It is instructive to compare it mentally with polynomial long division: the latter would require rewriting terms of the dividend multiple times and subtracting entire polynomials, whereas Ruffini reduces everything to a single row of scalar multiplications and additions. The compactness of the table is especially advantageous when the coefficients are numerically large.
Finding the key value
The divisor is \(x - 4\), already in standard form \(x - a\). The value that makes the divisor zero is: \[ x - 4 = 0 \;\Rightarrow\; x = 4 \] Therefore: \[ a = 4 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 - 5x^2 - 2x + 24 \] The corresponding coefficients are: \[ 1 \quad -5 \quad -2 \quad 24 \]
Setting up the synthetic division table
Place the value \(a = 4\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} 4 & 1 & -5 & -2 & 24 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = 4\) and place the result beneath the next coefficient:
\[ 1 \cdot 4 = 4 \]
Step 3: add the values in the column:
\[ -5 + 4 = -1 \]
Step 4: repeat the process: multiply the value obtained by \(4\) and add:
\[ -1 \cdot 4 = -4 \] \[ -2 + (-4) = -6 \]
Step 5: carry out the final step:
\[ -6 \cdot 4 = -24 \] \[ 24 + (-24) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 4 & 1 & -5 & -2 & 24 \\ & & 4 & -4 & -24 \\ \hline & 1 & -1 & -6 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 - x - 6 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 - x - 6} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 4\) and can be written as: \[ x^3 - 5x^2 - 2x + 24 = (x - 4)(x^2 - x - 6) \]
Exercise 9:
\[ (x^3 + 3x^2 - 10x - 24) : (x + 4) \]
Result
\[ x^2 - x - 6 \]
Solution
Approach
The divisor \(x + 4\) has a positive constant term, just like the cases \(x + 3\) and \(x + 2\) encountered earlier. Ruffini's rule applies in exactly the same way, but it is worth reiterating the key point: one does not enter \(+4\) in the table, but rather the root of the divisor, namely \(x = -4\). Confusing the constant term of the divisor with its root is the single most common mistake when applying this technique.
Finding the key value
The divisor is \(x + 4 = x - (-4)\). The value that makes the divisor zero is: \[ x + 4 = 0 \;\Rightarrow\; x = -4 \] Therefore: \[ a = -4 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 + 3x^2 - 10x - 24 \] The corresponding coefficients are: \[ 1 \quad 3 \quad -10 \quad -24 \]
Setting up the synthetic division table
Place the value \(a = -4\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} -4 & 1 & 3 & -10 & -24 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = -4\) and place the result beneath the next coefficient:
\[ 1 \cdot (-4) = -4 \]
Step 3: add the values in the column:
\[ 3 + (-4) = -1 \]
Step 4: repeat the process: multiply the value obtained by \(-4\) and add:
\[ -1 \cdot (-4) = 4 \] \[ -10 + 4 = -6 \]
Step 5: carry out the final step:
\[ -6 \cdot (-4) = 24 \] \[ -24 + 24 = 0 \]
Complete table
\[ \begin{array}{r|rrrr} -4 & 1 & 3 & -10 & -24 \\ & & -4 & 4 & 24 \\ \hline & 1 & -1 & -6 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 - x - 6 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 - x - 6} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x + 4\) and can be written as: \[ x^3 + 3x^2 - 10x - 24 = (x + 4)(x^2 - x - 6) \]
Exercise 10:
\[ (2x^3 + 3x^2 - 11x - 6) : (x - 2) \]
Result
\[ 2x^2 + 7x + 3 \]
Solution
Approach
The divisor is of the form \(x - a\), so Ruffini's rule applies. The dividend has leading coefficient \(2\): it is worth observing that this value passes through unchanged as the first entry of the bottom row, thereby determining the leading coefficient of the quotient. In other words, the quotient of a polynomial with leading coefficient \(k\) divided by a monic binomial will itself have leading coefficient \(k\).
Finding the key value
The divisor is \(x - 2\), already in standard form \(x - a\). The value that makes the divisor zero is: \[ x - 2 = 0 \;\Rightarrow\; x = 2 \] Therefore: \[ a = 2 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ 2x^3 + 3x^2 - 11x - 6 \] The corresponding coefficients are: \[ 2 \quad 3 \quad -11 \quad -6 \]
Setting up the synthetic division table
Place the value \(a = 2\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} 2 & 2 & 3 & -11 & -6 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 2 \]
Step 2: multiply the value just written by \(a = 2\) and place the result beneath the next coefficient:
\[ 2 \cdot 2 = 4 \]
Step 3: add the values in the column:
\[ 3 + 4 = 7 \]
Step 4: repeat the process: multiply the value obtained by \(2\) and add:
\[ 7 \cdot 2 = 14 \] \[ -11 + 14 = 3 \]
Step 5: carry out the final step:
\[ 3 \cdot 2 = 6 \] \[ -6 + 6 = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 2 & 2 & 3 & -11 & -6 \\ & & 4 & 14 & 6 \\ \hline & 2 & 7 & 3 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial. The leading coefficient \(2\) of the dividend appears unchanged as the first coefficient of the quotient: \[ 2x^2 + 7x + 3 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{2x^2 + 7x + 3} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 2\) and can be written as: \[ 2x^3 + 3x^2 - 11x - 6 = (x - 2)(2x^2 + 7x + 3) \]
Exercise 11:
\[ (x^3 - x^2 - 14x + 24) : (x - 3) \]
Result
\[ x^2 + 2x - 8 \]
Solution
Approach
Before applying Ruffini's rule, it is useful to recall the theorem on which it rests: the Remainder Theorem states that the remainder of the division of a polynomial \(p(x)\) by \((x - a)\) equals \(p(a)\). In particular, if \(p(a) = 0\) then the remainder is zero and \((x - a)\) is an exact divisor. Ruffini's rule is simply a compact, mechanical procedure for computing the quotient and remainder whose existence this theorem guarantees.
Finding the key value
The divisor is \(x - 3\), already in standard form \(x - a\). The value that makes the divisor zero is: \[ x - 3 = 0 \;\Rightarrow\; x = 3 \] Therefore: \[ a = 3 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 - x^2 - 14x + 24 \] The corresponding coefficients are: \[ 1 \quad -1 \quad -14 \quad 24 \]
Setting up the synthetic division table
Place the value \(a = 3\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} 3 & 1 & -1 & -14 & 24 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = 3\) and place the result beneath the next coefficient:
\[ 1 \cdot 3 = 3 \]
Step 3: add the values in the column:
\[ -1 + 3 = 2 \]
Step 4: repeat the process: multiply the value obtained by \(3\) and add:
\[ 2 \cdot 3 = 6 \] \[ -14 + 6 = -8 \]
Step 5: carry out the final step:
\[ -8 \cdot 3 = -24 \] \[ 24 + (-24) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 3 & 1 & -1 & -14 & 24 \\ & & 3 & 6 & -24 \\ \hline & 1 & 2 & -8 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 + 2x - 8 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 + 2x - 8} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 3\) and can be written as: \[ x^3 - x^2 - 14x + 24 = (x - 3)(x^2 + 2x - 8) \]
Exercise 12:
\[ (x^3 - 4x^2 - 7x + 10) : (x + 2) \]
Result
\[ x^2 - 6x + 5 \]
Solution
Approach
We apply Ruffini's rule and, once finished, verify the result in the most direct way possible: by expanding the product \((x + 2)(x^2 - 6x + 5)\) and checking that we recover the original polynomial. This verification habit — quick to carry out — allows one to catch any arithmetic errors made during the table immediately.
Finding the key value
The divisor is \(x + 2 = x - (-2)\). The value that makes the divisor zero is: \[ x + 2 = 0 \;\Rightarrow\; x = -2 \] Therefore: \[ a = -2 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 - 4x^2 - 7x + 10 \] The corresponding coefficients are: \[ 1 \quad -4 \quad -7 \quad 10 \]
Setting up the synthetic division table
Place the value \(a = -2\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} -2 & 1 & -4 & -7 & 10 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = -2\) and place the result beneath the next coefficient:
\[ 1 \cdot (-2) = -2 \]
Step 3: add the values in the column:
\[ -4 + (-2) = -6 \]
Step 4: repeat the process: multiply the value obtained by \(-2\) and add:
\[ -6 \cdot (-2) = 12 \] \[ -7 + 12 = 5 \]
Step 5: carry out the final step:
\[ 5 \cdot (-2) = -10 \] \[ 10 + (-10) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} -2 & 1 & -4 & -7 & 10 \\ & & -2 & 12 & -10 \\ \hline & 1 & -6 & 5 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 - 6x + 5 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 - 6x + 5} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x + 2\) and can be written as: \[ x^3 - 4x^2 - 7x + 10 = (x + 2)(x^2 - 6x + 5) \]
Exercise 13:
\[ (3x^3 - 5x^2 - 4x + 4) : (x - 2) \]
Result
\[ 3x^2 + x - 2 \]
Solution
Approach
Here again the divisor is of the form \(x - a\), so we apply Ruffini's rule. The leading coefficient of the dividend is \(3\): the intermediate products will all be multiples of \(3\), which does not increase the difficulty of the method but does require that no factor be overlooked. With non-unit leading coefficients it is good practice to double-check each multiplication before moving on to the next step.
Finding the key value
The divisor is \(x - 2\), already in standard form \(x - a\). The value that makes the divisor zero is: \[ x - 2 = 0 \;\Rightarrow\; x = 2 \] Therefore: \[ a = 2 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ 3x^3 - 5x^2 - 4x + 4 \] The corresponding coefficients are: \[ 3 \quad -5 \quad -4 \quad 4 \]
Setting up the synthetic division table
Place the value \(a = 2\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} 2 & 3 & -5 & -4 & 4 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 3 \]
Step 2: multiply the value just written by \(a = 2\) and place the result beneath the next coefficient:
\[ 3 \cdot 2 = 6 \]
Step 3: add the values in the column:
\[ -5 + 6 = 1 \]
Step 4: repeat the process: multiply the value obtained by \(2\) and add:
\[ 1 \cdot 2 = 2 \] \[ -4 + 2 = -2 \]
Step 5: carry out the final step:
\[ -2 \cdot 2 = -4 \] \[ 4 + (-4) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 2 & 3 & -5 & -4 & 4 \\ & & 6 & 2 & -4 \\ \hline & 3 & 1 & -2 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial. The leading coefficient \(3\) of the dividend reappears unchanged as the first coefficient of the quotient: \[ 3x^2 + x - 2 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{3x^2 + x - 2} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 2\) and can be written as: \[ 3x^3 - 5x^2 - 4x + 4 = (x - 2)(3x^2 + x - 2) \]
Exercise 14:
\[ (x^3 + 5x^2 + 2x - 8) : (x + 4) \]
Result
\[ x^2 + x - 2 \]
Solution
Approach
The divisor is \(x + 4\), with a constant term larger in absolute value than in the previous cases. In Ruffini's rule, this translates into larger intermediate products: it is advisable to carry out each multiplication carefully, since an error on a large value produces more conspicuous discrepancies in the final row. The procedure itself is unchanged: identify the root \(a = -4\) and proceed as usual.
Finding the key value
The divisor is \(x + 4 = x - (-4)\). The value that makes the divisor zero is: \[ x + 4 = 0 \;\Rightarrow\; x = -4 \] Therefore: \[ a = -4 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 + 5x^2 + 2x - 8 \] The corresponding coefficients are: \[ 1 \quad 5 \quad 2 \quad -8 \]
Setting up the synthetic division table
Place the value \(a = -4\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} -4 & 1 & 5 & 2 & -8 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = -4\) and place the result beneath the next coefficient:
\[ 1 \cdot (-4) = -4 \]
Step 3: add the values in the column:
\[ 5 + (-4) = 1 \]
Step 4: repeat the process: multiply the value obtained by \(-4\) and add:
\[ 1 \cdot (-4) = -4 \] \[ 2 + (-4) = -2 \]
Step 5: carry out the final step:
\[ -2 \cdot (-4) = 8 \] \[ -8 + 8 = 0 \]
Complete table
\[ \begin{array}{r|rrrr} -4 & 1 & 5 & 2 & -8 \\ & & -4 & -4 & 8 \\ \hline & 1 & 1 & -2 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 + x - 2 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 + x - 2} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x + 4\) and can be written as: \[ x^3 + 5x^2 + 2x - 8 = (x + 4)(x^2 + x - 2) \]
Exercise 15:
\[ (2x^3 - 9x^2 + 7x + 6) : (x - 3) \]
Result
\[ 2x^2 - 3x - 2 \]
Solution
Approach
Beyond yielding the quotient and remainder, Ruffini's rule is a factorization tool: whenever the remainder is zero, the polynomial can be written as the product of the divisor and the quotient. If the resulting quotient is itself further factorable — for instance by the quadratic formula — one obtains the complete factorization into irreducible factors of the original polynomial. In this exercise the quotient \(2x^2 - 3x - 2\) is a trinomial that can be factored further, but this lies beyond the current objective.
Finding the key value
The divisor is \(x - 3\), already in standard form \(x - a\). The value that makes the divisor zero is: \[ x - 3 = 0 \;\Rightarrow\; x = 3 \] Therefore: \[ a = 3 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ 2x^3 - 9x^2 + 7x + 6 \] The corresponding coefficients are: \[ 2 \quad -9 \quad 7 \quad 6 \]
Setting up the synthetic division table
Place the value \(a = 3\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} 3 & 2 & -9 & 7 & 6 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 2 \]
Step 2: multiply the value just written by \(a = 3\) and place the result beneath the next coefficient:
\[ 2 \cdot 3 = 6 \]
Step 3: add the values in the column:
\[ -9 + 6 = -3 \]
Step 4: repeat the process: multiply the value obtained by \(3\) and add:
\[ -3 \cdot 3 = -9 \] \[ 7 + (-9) = -2 \]
Step 5: carry out the final step:
\[ -2 \cdot 3 = -6 \] \[ 6 + (-6) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 3 & 2 & -9 & 7 & 6 \\ & & 6 & -9 & -6 \\ \hline & 2 & -3 & -2 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ 2x^2 - 3x - 2 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{2x^2 - 3x - 2} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 3\) and can be written as: \[ 2x^3 - 9x^2 + 7x + 6 = (x - 3)(2x^2 - 3x - 2) \]
Exercise 16:
\[ (x^3 - 7x + 6) : (x - 1) \]
Result
\[ x^2 + x - 6 \]
Solution
Approach
The dividend \(x^3 - 7x + 6\) contains no \(x^2\) term: it is a polynomial with a missing term. Before applying Ruffini's rule, it is essential to make the corresponding zero coefficient explicit by inserting \(0\) in the \(x^2\) position within the table. Omitting this step would cause every coefficient to be matched with the wrong power, invalidating the entire computation.
Finding the key value
The divisor is \(x - 1\), already in standard form \(x - a\). The value that makes the divisor zero is: \[ x - 1 = 0 \;\Rightarrow\; x = 1 \] Therefore: \[ a = 1 \]
Writing down the coefficients
The polynomial written in full with the missing term made explicit is: \[ x^3 + 0x^2 - 7x + 6 \] The coefficients in order are: \[ 1 \quad 0 \quad -7 \quad 6 \]
Setting up the synthetic division table
Place the value \(a = 1\) on the left and write all four coefficients — including the zero — along the top row:
\[ \begin{array}{r|rrrr} 1 & 1 & 0 & -7 & 6 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = 1\) and place the result beneath the next coefficient:
\[ 1 \cdot 1 = 1 \]
Step 3: add the values in the column:
\[ 0 + 1 = 1 \]
Step 4: repeat the process: multiply the value obtained by \(1\) and add:
\[ 1 \cdot 1 = 1 \] \[ -7 + 1 = -6 \]
Step 5: carry out the final step:
\[ -6 \cdot 1 = -6 \] \[ 6 + (-6) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 1 & 1 & 0 & -7 & 6 \\ & & 1 & 1 & -6 \\ \hline & 1 & 1 & -6 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 + x - 6 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 + x - 6} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 1\) and can be written as: \[ x^3 - 7x + 6 = (x - 1)(x^2 + x - 6) \]
Exercise 17:
\[ (x^3 - 2x^2 - 9x + 18) : (x + 3) \]
Result
\[ x^2 - 5x + 6 \]
Solution
Approach
Before setting up the table, it is worth quickly verifying that \(x + 3\) is indeed an exact divisor by computing \(p(-3)\) mentally: \((-3)^3 - 2(-3)^2 - 9(-3) + 18 = -27 - 18 + 27 + 18 = 0\). This preliminary check, made possible by the Remainder Theorem, takes only a few seconds and guarantees that the synthetic division table will yield a zero remainder. If the result were nonzero, we would already know the division is not exact, without needing to complete the entire table.
Finding the key value
The divisor is \(x + 3 = x - (-3)\). The value that makes the divisor zero is: \[ x + 3 = 0 \;\Rightarrow\; x = -3 \] Therefore: \[ a = -3 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 - 2x^2 - 9x + 18 \] The corresponding coefficients are: \[ 1 \quad -2 \quad -9 \quad 18 \]
Setting up the synthetic division table
Place the value \(a = -3\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} -3 & 1 & -2 & -9 & 18 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = -3\) and place the result beneath the next coefficient:
\[ 1 \cdot (-3) = -3 \]
Step 3: add the values in the column:
\[ -2 + (-3) = -5 \]
Step 4: repeat the process: multiply the value obtained by \(-3\) and add:
\[ -5 \cdot (-3) = 15 \] \[ -9 + 15 = 6 \]
Step 5: carry out the final step:
\[ 6 \cdot (-3) = -18 \] \[ 18 + (-18) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} -3 & 1 & -2 & -9 & 18 \\ & & -3 & 15 & -18 \\ \hline & 1 & -5 & 6 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 - 5x + 6 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact, as anticipated by the preliminary check.
Result
\[ \boxed{x^2 - 5x + 6} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x + 3\) and can be written as: \[ x^3 - 2x^2 - 9x + 18 = (x + 3)(x^2 - 5x + 6) \]
Exercise 18:
\[ (2x^3 + x^2 - 13x + 6) : (x + 3) \]
Result
\[ 2x^2 - 5x + 2 \]
Solution
Approach
This exercise combines two features already encountered separately: a divisor of the form \(x + k\) with \(k > 0\) — which requires extracting a negative root — and a leading coefficient of the dividend other than \(1\). Ruffini's rule handles both situations with no change to the procedure; the only difference is the extra care required when carrying out products that simultaneously involve negative numbers and integer coefficients greater than one.
Finding the key value
The divisor is \(x + 3 = x - (-3)\). The value that makes the divisor zero is: \[ x + 3 = 0 \;\Rightarrow\; x = -3 \] Therefore: \[ a = -3 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ 2x^3 + x^2 - 13x + 6 \] The corresponding coefficients are: \[ 2 \quad 1 \quad -13 \quad 6 \]
Setting up the synthetic division table
Place the value \(a = -3\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} -3 & 2 & 1 & -13 & 6 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 2 \]
Step 2: multiply the value just written by \(a = -3\) and place the result beneath the next coefficient:
\[ 2 \cdot (-3) = -6 \]
Step 3: add the values in the column:
\[ 1 + (-6) = -5 \]
Step 4: repeat the process: multiply the value obtained by \(-3\) and add:
\[ -5 \cdot (-3) = 15 \] \[ -13 + 15 = 2 \]
Step 5: carry out the final step:
\[ 2 \cdot (-3) = -6 \] \[ 6 + (-6) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} -3 & 2 & 1 & -13 & 6 \\ & & -6 & 15 & -6 \\ \hline & 2 & -5 & 2 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial. The leading coefficient \(2\) of the dividend reappears as the first coefficient of the quotient: \[ 2x^2 - 5x + 2 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{2x^2 - 5x + 2} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x + 3\) and can be written as: \[ 2x^3 + x^2 - 13x + 6 = (x + 3)(2x^2 - 5x + 2) \]
Exercise 19:
\[ (x^3 + x^2 - 5x + 3) : (x - 1) \]
Result
\[ x^2 + 2x - 3 \]
Solution
Approach
Applying Ruffini's rule does more than find the quotient and remainder: when the remainder is zero, \(a\) is a root of the dividend polynomial. In this exercise we will obtain the quotient \(x^2 + 2x - 3\), which is itself factorable as \((x - 1)(x + 3)\). This means that \(x = 1\) is a double root and \(x = -3\) is a further root, and the complete factorization of the original polynomial is \((x-1)^2(x+3)\). Ruffini's rule is thus the first step in a chain of factorizations.
Finding the key value
The divisor is \(x - 1\), already in standard form \(x - a\). The value that makes the divisor zero is: \[ x - 1 = 0 \;\Rightarrow\; x = 1 \] Therefore: \[ a = 1 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ x^3 + x^2 - 5x + 3 \] The corresponding coefficients are: \[ 1 \quad 1 \quad -5 \quad 3 \]
Setting up the synthetic division table
Place the value \(a = 1\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} 1 & 1 & 1 & -5 & 3 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 1 \]
Step 2: multiply the value just written by \(a = 1\) and place the result beneath the next coefficient:
\[ 1 \cdot 1 = 1 \]
Step 3: add the values in the column:
\[ 1 + 1 = 2 \]
Step 4: repeat the process: multiply the value obtained by \(1\) and add:
\[ 2 \cdot 1 = 2 \] \[ -5 + 2 = -3 \]
Step 5: carry out the final step:
\[ -3 \cdot 1 = -3 \] \[ 3 + (-3) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 1 & 1 & 1 & -5 & 3 \\ & & 1 & 2 & -3 \\ \hline & 1 & 2 & -3 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial: \[ x^2 + 2x - 3 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{x^2 + 2x - 3} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 1\) and can be written as: \[ x^3 + x^2 - 5x + 3 = (x - 1)(x^2 + 2x - 3) \] Factoring the quotient further gives the complete factorization: \[ x^3 + x^2 - 5x + 3 = (x - 1)^2(x + 3) \]
Exercise 20:
\[ (3x^3 + 2x^2 - 7x + 2) : (x - 1) \]
Result
\[ 3x^2 + 5x - 2 \]
Solution
Approach
Having reached the twentieth exercise, we can take stock of the method. Ruffini's rule has proved to be uniformly applicable regardless of the sign of the root, the absolute values of the coefficients, or the leading coefficient of the dividend. The only necessary condition remains that the divisor be a monic linear binomial \(x - a\): when this condition is satisfied, the three-row table always delivers the quotient and remainder quickly and reliably.
Finding the key value
The divisor is \(x - 1\), already in standard form \(x - a\). The value that makes the divisor zero is: \[ x - 1 = 0 \;\Rightarrow\; x = 1 \] Therefore: \[ a = 1 \]
Writing down the coefficients
The polynomial is already written in descending order of powers: \[ 3x^3 + 2x^2 - 7x + 2 \] The corresponding coefficients are: \[ 3 \quad 2 \quad -7 \quad 2 \]
Setting up the synthetic division table
Place the value \(a = 1\) on the left and write the coefficients along the top row:
\[ \begin{array}{r|rrrr} 1 & 3 & 2 & -7 & 2 \end{array} \]
Applying the rule step by step
Step 1: bring down the first coefficient to the bottom row unchanged.
\[ 3 \]
Step 2: multiply the value just written by \(a = 1\) and place the result beneath the next coefficient:
\[ 3 \cdot 1 = 3 \]
Step 3: add the values in the column:
\[ 2 + 3 = 5 \]
Step 4: repeat the process: multiply the value obtained by \(1\) and add:
\[ 5 \cdot 1 = 5 \] \[ -7 + 5 = -2 \]
Step 5: carry out the final step:
\[ -2 \cdot 1 = -2 \] \[ 2 + (-2) = 0 \]
Complete table
\[ \begin{array}{r|rrrr} 1 & 3 & 2 & -7 & 2 \\ & & 3 & 5 & -2 \\ \hline & 3 & 5 & -2 & 0 \end{array} \]
Reading the result
The values in the bottom row, excluding the last, represent the coefficients of the quotient polynomial. The leading coefficient \(3\) of the dividend appears unchanged as the first coefficient of the quotient: \[ 3x^2 + 5x - 2 \]
The last value is the remainder of the division: \[ r = 0 \] Since the remainder is zero, the division is exact.
Result
\[ \boxed{3x^2 + 5x - 2} \]
Conclusion
Since the remainder is zero, the original polynomial is divisible by \(x - 1\) and can be written as: \[ 3x^3 + 2x^2 - 7x + 2 = (x - 1)(3x^2 + 5x - 2) \]