Topic 1.1 · Unit 1 · CA CCSS A-APR.2, A-APR.6
Polynomial division, remainder and factor theorems
Dividing polynomials is how you break a cubic or quartic down to something you can factor. Synthetic division is much faster than long division, but only when the divisor is linear, knowing which tool applies is half the topic.
- Long division works for any divisor. Divide the leading terms, multiply back, subtract, bring down. The subtraction step is where signs go wrong, write the subtracted line with its signs already flipped.
- Synthetic division works only for divisors of the form \( x - c \). Write \( c \), then the coefficients. Bring down the first, multiply by \( c \), add, and repeat. The last number is the remainder; the rest are the quotient's coefficients, one degree lower.
- Insert zeros for missing degrees. \( x^3 - 8 \) has coefficients \( 1, 0, 0, -8 \). Skipping a missing term shifts everything and silently ruins the answer.
- Remainder theorem: the remainder on dividing \( P(x) \) by \( x - c \) is \( P(c) \). So evaluating is a one-line alternative to dividing when only the remainder is wanted.
- Factor theorem: \( x - c \) is a factor exactly when \( P(c) = 0 \). This is what turns "is 3 a root?" into a substitution rather than a division.
- Write the answer as \( \text{quotient} + \frac{\text{remainder}}{\text{divisor}} \).
Where marks are lost: the sign of \( c \). Dividing by \( x + 4 \) means \( c = -4 \), because \( x + 4 = x - (-4) \). Using \( +4 \) gives a wrong answer that still looks tidy.
Example 1: synthetic division. Divide \( 2x^3 - 5x^2 + 3x - 7 \) by \( x - 2 \).
Here \( c = 2 \) and the coefficients are \( 2, -5, 3, -7 \).
Bring down 2. Then \( 2 \times 2 = 4 \), \( -5 + 4 = -1 \). Then \( -1 \times 2 = -2 \), \( 3 - 2 = 1 \). Then \( 1 \times 2 = 2 \), \( -7 + 2 = -5 \).
Quotient \( 2x^2 - x + 1 \), remainder \( -5 \). So the answer is \( 2x^2 - x + 1 - \frac{5}{x-2} \).
Example 2: remainder theorem. Find the remainder when \( x^4 - 3x^2 + 5 \) is divided by \( x + 1 \).
\( x + 1 \) means \( c = -1 \). Evaluate: \( P(-1) = 1 - 3 + 5 = 3 \).
The remainder is 3, no division needed.
Example 3: factor theorem into a full factorization. Factor \( x^3 - 4x^2 + x + 6 \) given that \( x = 3 \) is a root.
Since \( P(3) = 27 - 36 + 3 + 6 = 0 \), \( x - 3 \) is a factor. Synthetic division with \( c = 3 \) on \( 1, -4, 1, 6 \) gives \( 1, -1, -2 \) with remainder 0.
So \( P(x) = (x-3)(x^2 - x - 2) = (x-3)(x-2)(x+1) \). Always try to factor the quotient further, stopping at the quadratic loses marks.
1–5 division, 6–10 the remainder and factor theorems, 11–15 full factorization and diagnosis.
- Divide \( x^2 + 5x + 6 \) by \( x + 2 \).
Show the full solution
\( c = -2 \) on \( 1, 5, 6 \): bring down 1; \( 5 - 2 = 3 \); \( 6 - 6 = 0 \). \( x + 3 \)
- Divide \( x^3 - 2x^2 - 5x + 6 \) by \( x - 1 \).
Show the full solution
\( c = 1 \) on \( 1, -2, -5, 6 \): \( 1, -1, -6, 0 \). \( x^2 - x - 6 \)
- Divide \( 3x^3 + 2x - 5 \) by \( x - 1 \).
Show the full solution
Insert the missing \( x^2 \): coefficients \( 3, 0, 2, -5 \). Synthetic with \( c = 1 \): \( 3, 3, 5, 0 \). \( 3x^2 + 3x + 5 \)
- Divide \( 2x^3 + 3x^2 - 1 \) by \( x + 2 \).
Show the full solution
\( c = -2 \) on \( 2, 3, 0, -1 \): \( 2, -1, 2, -5 \). \( 2x^2 - x + 2 - \frac{5}{x+2} \)
- Divide \( x^3 + 2x^2 - x - 2 \) by \( x^2 - 1 \) using long division.
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\( x^3 \div x^2 = x \); subtracting \( x^3 - x \) leaves \( 2x^2 - 2 \). Then \( 2x^2 \div x^2 = 2 \); subtracting \( 2x^2 - 2 \) leaves 0. \( x + 2 \)
- Find the remainder when \( x^3 - 4x + 1 \) is divided by \( x - 2 \).
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\( P(2) = 8 - 8 + 1 \). 1
- Find the remainder when \( 2x^3 + x^2 - 3 \) is divided by \( x + 1 \).
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\( c = -1 \): \( P(-1) = -2 + 1 - 3 = -4 \). \( -4 \)
- Is \( x - 3 \) a factor of \( x^3 - 2x^2 - 5x + 6 \)?
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\( P(3) = 27 - 18 - 15 + 6 = 0 \). Yes
- Is \( x + 2 \) a factor of \( x^3 + x^2 - 4x - 4 \)?
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\( P(-2) = -8 + 4 + 8 - 4 = 0 \). Yes
- Find \( k \) so that \( x - 2 \) is a factor of \( x^3 + kx - 6 \).
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Require \( P(2) = 0 \): \( 8 + 2k - 6 = 0 \), so \( 2k = -2 \). \( k = -1 \)
- Factor \( x^3 - 2x^2 - 5x + 6 \) completely.
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\( P(1) = 0 \), so divide by \( x - 1 \) to get \( x^2 - x - 6 \), which factors. \( (x-1)(x-3)(x+2) \)
- Factor \( x^3 + 3x^2 - 4x - 12 \) completely.
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Group: \( x^2(x+3) - 4(x+3) = (x+3)(x^2-4) \). \( (x+3)(x-2)(x+2) \)
- Solve \( x^3 - 7x + 6 = 0 \).
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\( P(1) = 0 \); dividing by \( x - 1 \) with coefficients \( 1, 0, -7, 6 \) gives \( x^2 + x - 6 = (x+3)(x-2) \). \( x = 1, 2, -3 \)
- A student divides \( x^3 - 8 \) by \( x - 2 \) using the coefficients \( 1, -8 \). Find
the error.
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The \( x^2 \) and \( x \) terms are missing and must be entered as zeros: \( 1, 0, 0, -8 \). Synthetic division with \( c = 2 \) then gives \( 1, 2, 4 \) remainder 0. \( x^2 + 2x + 4 \)
- A student divides by \( x + 5 \) using \( c = 5 \). Explain why that is wrong.
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Synthetic division uses the value that makes the divisor zero. \( x + 5 = 0 \) gives \( x = -5 \), so \( c = -5 \). Using \( +5 \) divides by \( x - 5 \) instead. \( c = -5 \)