Questions: Polynomial Long Division

Question 1

A student divides x³ + 5x − 3 by x² − 1. She writes the dividend as x³ + 5x − 3 without accounting for missing powers. What problem is most likely to follow?

• A. She will get a quotient of higher degree than possible
• B. The subtraction step will produce misaligned results because there is no x² term to subtract from
• C. She will stop the division process one step too early
• D. Nothing — missing terms can safely be ignored in polynomial long division

Answer: (B)

x³ + 5x − 3 is missing an x² term. Written correctly with the placeholder, it is x³ + 0x² + 5x − 3. When the first step produces a term in x², there is nothing to subtract it from if the placeholder was omitted — the columns fall out of alignment, corrupting every subsequent step. The placeholder 0x² is essential bookkeeping, not optional.

Question 2

When should you stop the polynomial long division process?

• A. After performing as many steps as the degree of the dividend
• B. When the quotient has the same degree as the divisor
• C. When the degree of the remainder is less than the degree of the divisor
• D. Only when the remainder is exactly zero

Answer: (C)

The process stops when the remaining expression (the current remainder) has degree strictly less than the divisor's degree — because the divisor can no longer 'fit into' it, just as 5 doesn't fit into 4 in integer division. You stop even if the remainder is nonzero. Stopping only when the remainder is zero would be incorrect; many divisions produce a nonzero remainder.

Question 3

If dividing f(x) by (x − a) produces a remainder of zero, then (x − a) is a factor of f(x).

• T. True
• F. False

Answer: True

By the polynomial division algorithm, f(x) = (x − a)·q(x) + r. If r = 0, then f(x) = (x − a)·q(x), which means (x − a) divides f(x) evenly — it is a factor. This connection is what the factor theorem formalizes and is one of the key reasons polynomial long division matters for factoring.

Question 4

In polynomial long division, you compare the degree of the remainder to the degree of the dividend to decide when to stop.

• T. True
• F. False

Answer: False

You compare the remainder's degree to the degree of the DIVISOR, not the dividend. The rule is: stop when deg(remainder) < deg(divisor). The dividend's degree is irrelevant to the stopping condition. Confusing divisor and dividend is a common source of errors in knowing when to halt the process.

Question 5

How does the equation f(x) = d(x)·q(x) + r(x) let you verify that a polynomial long division was performed correctly?

Model answer: Multiply the quotient q(x) by the divisor d(x), then add the remainder r(x). If the result equals the original dividend f(x), the division is correct. For example, if dividing x³ − 2x² − 4 by x − 2 gives quotient x² and remainder −4, verify by computing (x − 2)(x²) + (−4) = x³ − 2x² − 4, which matches the original.

This verification mirrors integer division: if 17 ÷ 5 gives quotient 3 and remainder 2, you check that 5·3 + 2 = 17. The same structure applies to polynomials. Multiplication is easier to perform than division and provides an independent check on the entire computation, catching sign errors, missed terms, and incorrect combining.