You want to use synthetic division to divide x³ - 2x² + 5 by x² - 1. What is the problem?
AYou cannot use synthetic division here because the divisor is not of the form x - c
BThe dividend has degree greater than 2, so synthetic division does not apply
CThere is no problem — synthetic division works for any divisor
DYou must first divide by x - 1, then divide again by x + 1
Synthetic division only works when the divisor is a linear polynomial of the exact form x - c (degree 1, leading coefficient 1). A quadratic divisor like x² - 1 requires polynomial long division. While x² - 1 = (x-1)(x+1) means you could divide by each factor separately, that is a multi-step process — not a direct application of synthetic division.
Question 2 Multiple Choice
If f(x) = x⁴ - 3x² + x - 7 and you divide f(x) by (x + 2), the Remainder Theorem says the remainder equals:
Af(2) = 16 - 12 + 2 - 7 = -1
Bf(-2) = 16 - 12 - 2 - 7 = -5
C0, because -2 must be a root
Df(0) = -7, the constant term
The Remainder Theorem states that dividing f(x) by (x - c) gives remainder f(c). Since the divisor is (x + 2) = (x - (-2)), we use c = -2. f(-2) = (-2)⁴ - 3(-2)² + (-2) - 7 = 16 - 12 - 2 - 7 = -5. Option A uses c = +2 — the classic sign error that occurs when students forget that dividing by (x + 2) means c = -2, not +2.
Question 3 True / False
If f(3) = 0, then (x - 3) is guaranteed to be a factor of f(x).
TTrue
FFalse
Answer: True
By the Remainder Theorem, the remainder when f(x) is divided by (x - 3) equals f(3). If f(3) = 0, the remainder is 0, which means (x - 3) divides f(x) evenly — it is a factor. This is the Factor Theorem, a direct consequence of the Remainder Theorem. It provides a powerful method for factoring: evaluate at candidate roots, and zeros produce linear factors.
Question 4 True / False
Synthetic division can be used to divide 2x³ - x + 4 by 2x - 3.
TTrue
FFalse
Answer: False
Standard synthetic division requires the divisor to have the form x - c: degree 1 and leading coefficient exactly 1. The divisor 2x - 3 has leading coefficient 2, not 1. Applying synthetic division without adjusting for the leading coefficient produces an incorrect quotient. You would need to use polynomial long division or factor out the 2 first.
Question 5 Short Answer
Explain why synthetic division is equivalent to evaluating the polynomial at x = c when the divisor is (x - c).
Think about your answer, then reveal below.
Model answer: By the Remainder Theorem, the remainder when f(x) is divided by (x - c) equals f(c). Synthetic division computes this remainder as its final entry. So running synthetic division with value c is a numerically efficient way to evaluate f(c) — essentially Horner's method for polynomial evaluation, which organizes the arithmetic to minimize operations.
This connection transforms synthetic division from a mere algebraic shortcut into a polynomial evaluation algorithm. For a degree-n polynomial, synthetic division evaluates f(c) in n multiplications and n additions, the same count as direct substitution but with better numerical organization. The Remainder Theorem is the bridge: dividing by (x - c) and evaluating at x = c are mathematically identical operations.