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Synthetic Substitution

Synthetic Substitution

The process of using synthetic division to evaluate p(c) for a polynomial p(x) and a number c.

Note: The remainder from synthetic division by x – c is equal to p(c).

 

 

See also

Remainder theorem

Key Formula

p(x)=anxn+an−1xn−1+⋯+a1x+a0⇒p(c)=remainder when dividing p(x) by (x−c)p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \quad \Rightarrow \quad p(c) = \text{remainder when dividing } p(x) \text{ by } (x - c)
Where:
  • p(x)p(x) = The polynomial you want to evaluate
  • an,an−1,…,a0a_n, a_{n-1}, \ldots, a_0 = The coefficients of the polynomial, listed from highest degree to lowest
  • cc = The specific number at which you evaluate the polynomial

Worked Example

Problem: Evaluate p(3) for p(x) = 2x³ − 5x² + 4x − 7 using synthetic substitution.
Step 1: Write down the value c = 3 and list the coefficients of p(x) in order of descending degree.
c=3Coefficients: 2,  −5,  4,  −7c = 3 \qquad \text{Coefficients: } 2, \; -5, \; 4, \; -7
Step 2: Bring the leading coefficient straight down. This starts the process.
Bring down 2\text{Bring down } 2
Step 3: Multiply the value you just brought down by c, then add the result to the next coefficient.
3×2=66+(−5)=13 \times 2 = 6 \qquad 6 + (-5) = 1
Step 4: Repeat: multiply the new result by c, then add to the next coefficient.
3×1=33+4=73 \times 1 = 3 \qquad 3 + 4 = 7
Step 5: Repeat once more for the last coefficient.
3×7=2121+(−7)=143 \times 7 = 21 \qquad 21 + (-7) = 14
Step 6: The final number is the remainder, which equals p(3).
p(3)=14p(3) = 14
Answer: p(3) = 14

Another Example

Problem: Evaluate p(−2) for p(x) = x⁴ + 0x³ − 3x² + x + 5 using synthetic substitution.
Step 1: Set c = −2 and list all coefficients, including a 0 for the missing x³ term.
c=−2Coefficients: 1,  0,  −3,  1,  5c = -2 \qquad \text{Coefficients: } 1, \; 0, \; -3, \; 1, \; 5
Step 2: Bring down the leading coefficient 1.
Bring down 1\text{Bring down } 1
Step 3: Multiply by −2 and add to the next coefficient.
(−2)(1)=−2−2+0=−2(-2)(1) = -2 \qquad -2 + 0 = -2
Step 4: Continue the process.
(−2)(−2)=44+(−3)=1(-2)(-2) = 4 \qquad 4 + (-3) = 1
Step 5: Continue.
(−2)(1)=−2−2+1=−1(-2)(1) = -2 \qquad -2 + 1 = -1
Step 6: Final step gives the remainder.
(−2)(−1)=22+5=7(-2)(-1) = 2 \qquad 2 + 5 = 7
Answer: p(−2) = 7

Frequently Asked Questions

What is the difference between synthetic substitution and synthetic division?
They use the exact same mechanical process. The difference is the purpose: synthetic division focuses on finding the quotient polynomial and remainder when dividing by (x − c), while synthetic substitution focuses specifically on that remainder because it equals p(c). Think of synthetic substitution as synthetic division done for the purpose of evaluation.
Why is synthetic substitution faster than direct substitution?
Direct substitution requires computing each power of c separately and then multiplying and adding, which can involve large intermediate calculations. Synthetic substitution reduces the work to a sequence of single multiplications and additions — one pair per coefficient — making it faster and less error-prone, especially for high-degree polynomials.

Synthetic Substitution vs. Direct Substitution

Direct substitution means plugging c into p(x) and computing each term (e.g., raising c to each power, multiplying by coefficients, then summing). Synthetic substitution achieves the same result — finding p(c) — but uses the streamlined synthetic division algorithm, which requires only repeated multiplication and addition. Both give identical answers; synthetic substitution is generally faster for polynomials of degree 3 or higher.

Why It Matters

Synthetic substitution is one of the quickest ways to evaluate a polynomial at a given value, which comes up constantly when testing possible rational zeros of a polynomial. It also reinforces the Remainder Theorem, connecting polynomial evaluation directly to division. Mastering this technique saves significant time on algebra exams and standardized tests.

Common Mistakes

Mistake: Forgetting to include a 0 coefficient for any missing degree term.
Correction: Before starting, write out every coefficient from the highest degree down to the constant. If a power of x is missing (e.g., no x³ term in a degree-4 polynomial), insert 0 as its coefficient. Skipping this shifts all subsequent calculations and produces a wrong answer.
Mistake: Using the wrong sign for c when the divisor is (x + c).
Correction: Remember that synthetic division always uses the value c that makes the divisor equal zero. For (x + 3), you use c = −3, not 3. The divisor (x − c) means c is the number you substitute.

Related Terms