You want to divide p(x) = x³ + 2x² − 5x + 1 by (x + 4). What value of c should you write in the box for synthetic division?
A4
B-4
C-5
D1
Synthetic division divides by (x − c), so you use the value of c — the opposite of the sign in the divisor. Since x + 4 = x − (−4), c = −4. Using c = 4 is the most common error in synthetic division. The pattern to remember: when the divisor is (x + k), c = −k. Getting the sign wrong flips every multiplication step and produces a completely incorrect result.
Question 2 Multiple Choice
After performing synthetic division of p(x) by (x − 3), the last number in the synthetic division row is 7. What does this tell you about p(3)?
Ap(3) = 0, because the remainder represents the root at x = 3
Bp(3) = 7, by the Remainder Theorem — the remainder equals the polynomial evaluated at x = c
Cp(3) = 3, because c = 3 was used in the division
Dp(3) cannot be determined from the remainder alone
The Remainder Theorem states that when p(x) is divided by (x − c), the remainder equals p(c). So a remainder of 7 means p(3) = 7. Synthetic division simultaneously divides the polynomial and evaluates it at x = c. If the remainder were 0, that would mean x = 3 is a root. This connection makes synthetic division the computational engine for testing potential roots via the Factor Theorem.
Question 3 True / False
When using synthetic division to divide a degree-4 polynomial by (x − 2), you must write exactly 5 numbers in the coefficient row — one for each degree from x⁴ down to x⁰, inserting 0 for any missing terms.
TTrue
FFalse
Answer: True
Every degree position must be represented, even if its coefficient is zero. If the polynomial has no x² term, you must write 0 in that slot. Skipping a missing term shifts all remaining coefficients one position to the left, making every subsequent multiplication and addition wrong and producing a quotient with incorrectly assigned degrees.
Question 4 True / False
Synthetic division can be used to divide a polynomial by any divisor, including quadratics like (x² − 3x + 2).
TTrue
FFalse
Answer: False
Synthetic division only works for linear divisors of the form (x − c). The algorithm's multiply-and-add pattern relies on the divisor having exactly one degree. For divisors of degree 2 or higher, the pattern breaks down and produces meaningless results; polynomial long division must be used instead. This restriction is the most important limitation to remember when choosing which method to apply.
Question 5 Short Answer
Explain why the remainder in synthetic division equals p(c), and why this makes synthetic division more than just a computational shortcut.
Think about your answer, then reveal below.
Model answer: When p(x) is divided by (x − c), there exist quotient q(x) and remainder r such that p(x) = (x − c) · q(x) + r. Substituting x = c gives p(c) = (c − c) · q(c) + r = 0 + r = r. So the remainder must equal p(c). This means every synthetic division simultaneously divides the polynomial and evaluates it at x = c — making it a fast way to test whether c is a root (remainder = 0), which is the computational foundation of the Factor Theorem and Rational Root Theorem.
This algebraic connection is what elevates synthetic division from arithmetic shortcut to root-finding tool. Every time you perform synthetic division, you are implicitly answering 'does this polynomial equal zero at x = c?' without needing a separate evaluation step. Understanding this connection — rather than just memorizing the bring-down-multiply-add procedure — makes the method useful in the broader context of polynomial analysis.