DWG NO. 2.3 — Lesson 3 of 6

Polynomial Division and the Remainder/Factor Theorems

Unit 2: Polynomial and Rational Functions · ~25 min

Objective Divide a polynomial by a linear binomial using long division or synthetic division, and apply the Remainder and Factor Theorems.

Long division of polynomials

Dividing polynomials works the same way long division of numbers does: divide the leading terms, multiply that result by the whole divisor, subtract, bring down the next term, and repeat until nothing's left to bring down. Whatever's left over at the end is the remainder.

Synthetic division — the shortcut

When the divisor has the simple form (x − c), there's a faster path: write down just the coefficients of the dividend, bring down the first one, then repeatedly multiply by c and add down the next column. It gives the same quotient and remainder as long division, with far less writing.

The Remainder Theorem

Dividing f(x) by (x − c) always produces f(x) = (x−c)·q(x) + r for some quotient q(x) and remainder r. Substitute x = c into that equation: every term with (x−c) disappears, leaving f(c) = r. In other words: the remainder is the same number you'd get by evaluating f(c) directly — synthetic division becomes a way to evaluate a polynomial at a value without substituting by hand.

The Factor Theorem

A direct consequence: if the remainder r is 0, then f(x) = (x−c)·q(x) exactly, with nothing left over — so (x−c) is a factor of f(x). And since r = f(c), that's the same as saying f(c) = 0. This gives a two-way test: (x−c) is a factor of f(x) if and only if f(c) = 0.

f(x) = (x−c)·q(x) + r — the remainder from division is always f(c)

Worked Example 1 · Long division

ProblemDivide f(x) = 2x³ − 3x² + 4x − 5 by (x − 2).

12x³ ÷ x = 2x². Multiply: 2x²(x−2) = 2x³−4x². Subtract: −3x²−(−4x²) = x². Bring down 4x: x²+4x.

2x² ÷ x = x. Multiply: x(x−2) = x²−2x. Subtract: 4x−(−2x) = 6x. Bring down −5: 6x−5.

36x ÷ x = 6. Multiply: 6(x−2) = 6x−12. Subtract: −5−(−12) = 7.

Quotient: 2x² + x + 6 · Remainder: 7

Worked Example 2 · Synthetic division confirming a factor

ProblemDivide f(x) = x³ + 4x² − 5x − 14 by (x − 2) using synthetic division with c = 2 and coefficients 1, 4, −5, −14.

1Bring down 1. Multiply by 2, add: 4+2=6. Multiply by 2, add: −5+12=7. Multiply by 2, add: −14+14=0.

2The remainder is 0, so by the Factor Theorem, (x−2) is a factor of f(x) — and f(2) must equal 0.

Quotient: x² + 6x + 7 · Remainder: 0 — so (x−2) is a factor

Worked Example 3 · Using the Remainder Theorem to evaluate f(c)

ProblemFind f(−3) for f(x) = x³ − 2x² − 5x + 6 using synthetic division instead of direct substitution.

1c = −3, coefficients 1, −2, −5, 6. Bring down 1. Multiply by −3, add: −2+(−3)=−5. Multiply by −3, add: −5+15=10. Multiply by −3, add: 6+(−30)=−24.

2The remainder is −24, so by the Remainder Theorem, f(−3) = −24.

f(−3) = −24

Guided practice

1.Divide f(x) = 3x² + 5x − 2 by (x + 2) using synthetic division. State the quotient and remainder.
Show answer

c = −2, coefficients 3, 5, −2. Bring down 3. 3(−2)+5=−1. −1(−2)+(−2)=0. Quotient: 3x − 1 · Remainder: 0
2.Use the Remainder Theorem (via synthetic division) to find f(3) for f(x) = x³ − 4x² + x + 6.
Show answer

c = 3, coefficients 1, −4, 1, 6. 1(3)+(−4)=−1. −1(3)+1=−2. −2(3)+6=0. f(3) = 0
3.Is (x − 1) a factor of f(x) = x³ − 6x² + 11x − 6? Use the Factor Theorem to decide.
Show answer

f(1) = 1 − 6 + 11 − 6 = 0. Since f(1) = 0, yes, (x−1) is a factor.
4.Divide f(x) = 2x³ − x² − 7x + 2 by (x − 1). State the quotient and remainder.
Show answer

c = 1, coefficients 2, −1, −7, 2. 2(1)+(−1)=1. 1(1)+(−7)=−6. −6(1)+2=−4. Quotient: 2x² + x − 6 · Remainder: −4
5.Dividing f(x) by (x + 5) gives a remainder of 0. What does that tell you about f(−5), and about the factors of f(x)?
Show answer

By the Remainder Theorem, f(−5) = 0. By the Factor Theorem, since the remainder is 0, (x+5) must be a factor of f(x).

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