Synthetic Division When the Divisor Is ax - b

You have a polynomial to divide by something like 2x + 1, and the synthetic division with leading coefficient method you memorised only works for x − c. The fix takes two extra seconds: set c = b / a, run the table, then divide every coefficient in the quotient row by a. That is the whole adjustment. The remainder stays untouched.

The standard synthetic division algorithm, as defined in OpenStax College Algebra 2e section 5.4 and every algebra textbook since Ruffini published the rule in 1809, assumes a divisor of the form x − k. That assumption is why the method uses only the coefficients and a single value c. When the divisor is 2x − 1, you cannot feed a 2 into the standard table and get a correct quotient out. The algorithm has no slot for the leading coefficient of the divisor. The table was built for monic linear divisors, and it will produce wrong numbers if you try to force it.

You keep the synthetic division shortcut and avoid long division by transforming the divisor first. For 2x − 1, rewrite it as x − (1/2). That is the step that matters. Use c = 1/2 in the table. The bottom row you get is a set of quotient coefficients that are off by a factor of 2. Divide each of those coefficients by 2, and you have the real quotient. The remainder in the bottom row is the actual remainder, no division needed there. Paul's Online Math Notes shows the same procedure for ax − b: use the root b/a, then rescale the quotient coefficients by a.

Two cases come up: a positive a and a negative a. An alternative method factors a out of the divisor first. The fastest check is multiplying back. The mistake that trips most students is forgetting to rescale the quotient. Do not be that person. Divide the quotient coefficients by a.

Synthetic division is a coefficient-only algorithm. You write the coefficients of the dividend, choose c from x − c, and run the bring-down, multiply-and-add loop. The divisor itself never appears in the table. That is the source of both its speed and its limitation.

When the divisor is x − 2, c = 2. When the divisor is x + 5, c = −5. That sign flip is the most common error in all of synthetic division, and it happens because the algorithm is built around the root of the divisor, not the sign you see written. The divisor x − c has root c. The divisor x + 5 has root −5. If you set c = +5 for x + 5, every number in your table is wrong.

The algorithm cannot absorb a coefficient on x. The divisor 2x − 1 has a root of 1/2, but the table treats that root as if it came from x − (1/2). The quotient coefficients that drop out of the table are the ones you would get if you divided by x − (1/2). But you did not divide by x − (1/2). You divided by 2x − 1, which is x − (1/2) multiplied by 2. Your quotient must be divided by 2 to correct for that missing factor.

If you skip the division step, your quotient will be exactly a times too large.

The procedure has four steps. First, write the divisor in the form ax − b. Identify a and b. For 2x − 1, a = 2 and b = 1. For 3x + 6, rewrite as 3x − (−6), so a = 3 and b = −6. Second, set c = b / a. For 2x − 1, c = 1/2. For 3x + 6, c = −6 / 3 = −2. Third, run synthetic division using that c exactly as you would for any x − c divisor. Bring down the leading coefficient. Multiply by c, add to the next coefficient, repeat. The last number in the bottom row is the remainder, and it is already correct for the original divisor ax − b.

Fourth, divide each coefficient in the bottom row except the last one by a. Those rescaled numbers are the coefficients of the true quotient polynomial. The degree of the quotient is one less than the degree of the dividend, as it always is for a linear divisor. The remainder stays exactly as it came out of the table.

Step-by-Step Table for (4x³ − 2x² + 6x − 1) ÷ (2x − 1)

Take the example (4x³ − 2x² + 6x − 1) ÷ (2x − 1). Write the dividend coefficients: 4, −2, 6, −1. Set c = 1/2. Run the table:

Bring down 4. Multiply 4 by 1/2 = 2. Add to −2: 0. Multiply 0 by 1/2 = 0. Add to 6: 6. Multiply 6 by 1/2 = 3. Add to −1: 2. The bottom row is 4, 0, 6, 2. The last number 2 is the remainder. The quotient coefficients before rescaling are 4, 0, 6. Divide each by a = 2: 2, 0, 3. The quotient is 2x² + 0x + 3, or just 2x² + 3. The remainder is 2.

Check by the Remainder Theorem: evaluate P(1/2) = 4(1/8) − 2(1/4) + 6(1/2) − 1 = 0.5 − 0.5 + 3 − 1 = 2. Matches the remainder from the table. The quotient 2x² + 3 multiplied by (2x − 1) gives back 4x³ − 2x² + 6x − 1 plus the remainder? Actually, (2x − 1)(2x² + 3) = 4x³ + 6x − 2x² − 3 = 4x³ − 2x² + 6x − 3. Add the remainder 2: 4x³ − 2x² + 6x − 1. Correct.

This is the case every textbook uses, and it is the one you will see first on a test. The divisor is 2x − 1, so a = 2 and b = 1. c = 1/2. Write the coefficients of the dividend: 4, −2, 6, −1. Set up the synthetic division table with c = 1/2 on the left.

Bring down the leading coefficient 4. Multiply 4 × 1/2 = 2. Add 2 to the next coefficient −2: result 0. Multiply 0 × 1/2 = 0. Add 0 to 6: result 6. Multiply 6 × 1/2 = 3. Add 3 to −1: result 2.

Bottom row: 4, 0, 6, 2. The last number, 2, is the remainder. The quotient coefficients before rescaling are 4, 0, 6. Divide each by a = 2: 2, 0, 3. The quotient polynomial is 2x² + 0x + 3, or 2x² + 3.

Common mistake: a student writes the quotient as 4x² + 0x + 6, forgetting to divide by a. That is the single most frequent error in synthetic division with a non‑monic divisor. The remainder 2 is correct either way, so the student might think the whole answer is right. Check by multiplying back: (2x − 1)(4x² + 6) = 8x³ + 12x − 4x² − 6, plus remainder 2 gives 8x³ − 4x² + 12x − 4. That does not match the original dividend. The rescale step is not optional.

Negative a changes nothing except the sign of c and the divisor you divide by at the end. Take (−3x³ + 5x² − x + 2) ÷ (−2x + 4). Rewrite the divisor as −2x + 4. That is of the form ax − b with a = −2 and b = −4 because (−2x + 4) = (−2)x − (−4). c = b / a = (−4) / (−2) = 2.

Dividend coefficients: −3, 5, −1, 2. Bring down −3. Multiply −3 × 2 = −6. Add to 5: −1. Multiply −1 × 2 = −2. Add to −1: −3. Multiply −3 × 2 = −6. Add to 2: −4. Bottom row: −3, −1, −3, −4. The remainder is −4.

Quotient coefficients before rescaling: −3, −1, −3. Divide each by a = −2: 1.5, 0.5, 1.5. The quotient is 1.5x² + 0.5x + 1.5, which you can write as (3/2)x² + (1/2)x + 3/2. Multiply back to check: (−2x + 4)(1.5x² + 0.5x + 1.5) = −3x³ − x² − 3x + 6x² + 2x + 6 = −3x³ + 5x² − x + 6. Add the remainder −4: −3x³ + 5x² − x + 2. Matches the dividend.

The sign of a does not change the algorithm. c is b / a. If a is negative, c might be positive or negative depending on b. The division of quotient coefficients by a will produce fractions if necessary. That is fine. The Rational Zero Theorem still applies if you are checking for rational roots, and the Factor Theorem works on the transformed divisor without change.

You can also handle a non‑monic divisor by factoring a out entirely before starting synthetic division. Write ax − b as a(x − b/a). Divide the original polynomial by (x − b/a) using standard synthetic division, then divide the whole quotient by a at the end.

For (4x³ − 2x² + 6x − 1) ÷ (2x − 1), factor 2 out of the divisor: 2(x − 1/2). Divide the polynomial by (x − 1/2) using synthetic division with c = 1/2. The bottom row gives quotient coefficients 4, 0, 6 and remainder 2. Now divide that quotient by the factor you pulled out: divide 4x² + 0x + 6 by 2, getting 2x² + 0x + 3. Remainder is still 2.

This method is mathematically identical to the c = b/a approach. Some students find it more intuitive because it separates the factoring step from the synthetic division step. The risk is the same: you must remember to divide the quotient by a. The remainder is never divided. If you forget the division, the quotient is wrong by the same factor a.

Use whichever method keeps the sign straight for you. Both produce the same result. The key is that synthetic division itself never handles the leading coefficient of the divisor, that coefficient is handled before or after the table, not inside it.

The fastest check is to multiply the quotient you got by the original divisor, add the remainder, and confirm the result equals the original dividend. This takes about thirty seconds and catches every common mistake: wrong c, missing zero coefficients, and the rescaling error.

For (4x³ − 2x² + 6x − 1) ÷ (2x − 1), the correct quotient is 2x² + 0x + 3. Multiply (2x − 1)(2x² + 3) = 4x³ + 6x − 2x² − 3 = 4x³ − 2x² + 6x − 3. Add remainder 2: 4x³ − 2x² + 6x − 1. Matches.

If you forgot to divide the quotient by a, you would multiply (2x − 1)(4x² + 6) = 8x³ + 12x − 4x² − 6. Add remainder 2: 8x³ − 4x² + 12x − 4. That does not match 4x³ − 2x² + 6x − 1, and the mismatch tells you exactly where the error is: the quotient coefficients are too large by a factor of 2.

For the negative a example, (−2x + 4)(1.5x² + 0.5x + 1.5) = −3x³ + 5x² − x + 6. Add remainder −4: −3x³ + 5x² − x + 2. Matches. If you had forgotten to rescale, the check would fail, and the failure points back to the quotient.

Do not skip the check. It is the only way to be sure you did not make an arithmetic error in the multiply-and-add loop, which is the second most common failure mode after the sign error for c.

For synthetic division with divisor 2x + 1, rewrite it as 2x − (−1). a = 2, b = −1. c = b / a = −1 / 2 = −0.5. Run synthetic division with c = −0.5. Take the quotient coefficients from the bottom row (excluding the last number) and divide each by 2. The last number in the bottom row is the remainder, unchanged.

Example: (6x² + 5x − 4) ÷ (2x + 1). c = −0.5. Coefficients: 6, 5, −4. Bring down 6. Multiply 6 × (−0.5) = −3. Add to 5: 2. Multiply 2 × (−0.5) = −1. Add to −4: −5. Bottom row: 6, 2, −5. Remainder is −5. Quotient coefficients before rescaling: 6, 2. Divide by 2: 3, 1. Quotient: 3x + 1. Check: (2x + 1)(3x + 1) = 6x² + 2x + 3x + 1 = 6x² + 5x + 1. Add remainder −5: 6x² + 5x − 4. Correct.

Forgetting to divide the quotient by a here would give 6x + 2, which multiplied back gives 12x² + 10x + 2, plus remainder −5 gives 12x² + 10x − 3. Does not match. The rescale step is the one thing that separates a correct answer in synthetic division with a non‑monic divisor from a wrong one.

For any divisor of the form ax − b, the procedure is the same. Identify a and b. Compute c = b / a. Run synthetic division with c. Divide the quotient coefficients by a. The remainder is untouched. This works for positive a, negative a, fractional a, and a = 1 (where it reduces to the standard case).

When a = 1, the divisor is x − b, and dividing the quotient coefficients by 1 does nothing. That is why the standard synthetic division algorithm never mentions the rescale step, it is invisible when a = 1. The moment a deviates from 1, the rescale step becomes necessary, and textbooks that omit it leave students stuck.

OpenStax College Algebra 2e section 5.4 states that synthetic division is defined only for a divisor of the form x − k. For non‑monic divisors, the textbook notes that you must transform the divisor to x − (b/a) before applying synthetic division, then adjust the quotient. Paul's Online Math Notes is clearer: use the root b/a, then divide the quotient coefficients by a. The standard algebra text treatment across all sources agrees: the remainder from synthetic division with root b/a is the actual remainder, and only the quotient needs rescaling.

The failure rate for this problem on timed tests is disproportionately high because students are taught the standard algorithm, see a non‑monic divisor, and either freeze or force the wrong c into the table. If you force c = 1 into the table for 2x − 1, the table runs but produces nonsense. The only fix is to handle the leading coefficient explicitly.

When the divisor is written as ax + b, rewrite it as ax − (−b). That puts it into the ax − b form. For 3x + 5, a = 3 and b = −5. c = b / a = −5 / 3. For −4x + 7, a = −4 and b = −7 because (−4x + 7) = (−4)x − (−7). c = (−7) / (−4) = 7/4.

The sign confusion is worse here than in any other variant. Students see 3x + 5 and want to set c = −5, which is wrong. c is b / a, and b is the constant term after you have written the divisor as ax − b. For 3x + 5, that means b = −5, so c = −5/3. Setting c = −5 would be correct only if a = 1. With a = 3, c = −5/3 is not the same as −5.

Worked example: (2x² + 7x + 3) ÷ (3x + 5). a = 3, b = −5, c = −5/3. Dividend coefficients: 2, 7, 3. Bring down 2. Multiply 2 × (−5/3) = −10/3. Add to 7: 21/3 − 10/3 = 11/3. Multiply (11/3) × (−5/3) = −55/9. Add to 3: 27/9 − 55/9 = −28/9. Bottom row: 2, 11/3, −28/9. Remainder is −28/9. Quotient coefficients before rescaling: 2, 11/3. Divide by a = 3: 2/3, 11/9. Quotient: (2/3)x + 11/9. Check by multiplying back: (3x + 5)((2/3)x + 11/9) = 2x² + (11/3)x + (10/3)x + 55/9 = 2x² + (21/3)x + 55/9 = 2x² + 7x + 55/9. Add remainder −28/9: 2x² + 7x + 27/9 = 2x² + 7x + 3. Correct.