Polynomial division and remainders · Teacher reference · answers included

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Teacher reference: contains answers, conditions and proofs.

04 Polynomial division and remainders · g10-04-01 · 5–10 min · TEACHER · ANSWERS

What does division preserve?

How can the quotient and remainder reconstruct f?

Predict: For a cubic divided by x−a, predict the degree of the quotient and the shape of the remainder.

Compare	Computed observation
Nonzero remainder x² coefficient b = -2; x coefficient c = 0; Constant d = 3; Divisor x−a = 2	Division identity: f(x)=(x−(2))(x^2)+(3)
An exact factor x² coefficient b = 0; x coefficient c = -1; Constant d = 0; Divisor x−a = 1	Division identity: f(x)=(x−(1))(x^2 +x)+(0)
Divide by x x² coefficient b = 2; x coefficient c = -3; Constant d = 4; Divisor x−a = 0	Division identity: f(x)=(x−(0))(x^2 +2x −3)+(4)

Conclusion: Polynomial division gives f=(x−a)q+r, with constant r. The two plotted curves f and (x−a)q differ vertically by r everywhere. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: The divisor is the nonzero monic polynomial x−a; coefficients are real.

Justification: Expanding the computed quotient gives coefficient equality for every power. The remainder degree is below one, proving the identity for all real x.

Ask: Point to the constant vertical difference at several positions, then ask for the coefficient proof.

Watch for: The identity holds at a too; only writing f/(x−a)=q+r/(x−a) excludes a.

Extension: Explain why a nonconstant remainder would violate division by a linear polynomial.

Exit: Divide x²+1 by x−1.

Quotient x+1, remainder 2; x²+1=(x−1)(x+1)+2.

Before: Polynomial expansion · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

04 Polynomial division and remainders · g10-04-02 · 5–10 min · TEACHER · ANSWERS

Where did the missing terms go?

Can missing coefficients be omitted in synthetic division?

Predict: For x⁴+bx²+c, predict the five coefficient slots before dividing by x−a.

Compare	Computed observation
Two internal zero slots x² coefficient b = -3; Constant c = 2; Synthetic multiplier a = 1	Division identity: f(x)=(x−(1))(x^3 +x^2 −2x −2)+(0)
Only x⁴ remains x² coefficient b = 0; Constant c = 0; Synthetic multiplier a = 2	Division identity: f(x)=(x−(2))(x^3 +2x^2 +4x +8)+(16)
Zero multiplier x² coefficient b = 2; Constant c = -1; Synthetic multiplier a = 0	Division identity: f(x)=(x−(0))(x^3 +2x)+(-1)

Conclusion: Missing x³ and x terms have coefficient zero. Keeping their slots preserves the powers during multiply-and-add steps. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: The polynomial has degree four even when b or c is zero. Division is by x−a.

Justification: Horner recurrence h₀=1, hᵢ=cᵢ+a hᵢ₋₁ matches successive coefficient cancellation. With zero slots retained, expansion recovers the original coefficients.

Ask: Ask pupils to label the power above every coefficient before moving counters.

Watch for: Removing a zero slot changes the polynomial rather than simplifying its notation.

Extension: Compare the erroneous division using [1,b,c] with the correct five-slot computation.

Exit: List the coefficient row for x⁴−3x²+2.

[1,0,−3,0,2], including both missing terms.

Before: Polynomial coefficient order · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

04 Polynomial division and remainders · g10-04-03 · 5–10 min · TEACHER · ANSWERS

Read a remainder from one point

Why is the remainder f(a)?

Predict: Divide f(x)=x³+bx+c by x−a; predict which graph input reveals the remainder.

Compare	Computed observation
Positive a x coefficient b = -2; Constant c = 3; Divisor zero a = 2	Division identity: f(x)=(x−(2))(x^2 +2x +2)+(7)
Divisor x+2 x coefficient b = 0; Constant c = 1; Divisor zero a = -2	Division identity: f(x)=(x−(-2))(x^2 −2x +4)+(-7)
Zero remainder x coefficient b = -1; Constant c = 0; Divisor zero a = 1	Division identity: f(x)=(x−(1))(x^2 +x)+(0)

Conclusion: At x=a the quotient term is multiplied by zero, leaving f(a)=r. The sign of a comes from the complete divisor. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: The divisor must be x−a; division by x+a uses input −a.

Justification: The division identity f=(x−a)q+r is a polynomial identity, so substitution x=a is valid and gives r=f(a) exactly.

Ask: Cover the long-division steps and ask why the marked ordinate is sufficient.

Watch for: A point with small output is not necessarily an exact zero.

Extension: Recover the remainders for x−2 and x+2 without performing long division.

Exit: Find the remainder of x³+1 divided by x+2.

f(−2)=−8+1=−7.

Before: Division identity and substitution · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

04 Polynomial division and remainders · g10-04-04 · 5–10 min · TEACHER · ANSWERS

Design a factor by changing one coefficient

What constant makes x−a a factor?

Predict: For x³+kx+c, predict the required c before comparing the original and repaired graphs.

Compare	Computed observation
A repair is needed Required root a = 1; x coefficient k = 2; Original constant c = 0	Required c=-3; currently a factor: false.
Already a factor Required root a = -1; x coefficient k = 2; Original constant c = 3	Required c=3; currently a factor: true.
Root at zero Required root a = 0; x coefficient k = -2; Original constant c = 0	Required c=0; currently a factor: true.

Conclusion: x−a is a factor iff f(a)=0, so c=−a³−ka. Adjusting c translates the graph vertically. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: The leading coefficient stays one; only the constant is repaired.

Justification: Remainder theorem gives r=a³+ka+c. The divisor is a factor exactly when r=0, which has the unique displayed solution for c.

Ask: Ask pupils to justify both directions of “if and only if”.

Watch for: A factor condition is an exact equality, not merely an axis crossing nearby.

Extension: If two different roots are required, determine whether one adjustable constant is sufficient.

Exit: Find c so x+1 divides x³+2x+c.

Set f(−1)=−1−2+c=0, giving c=3.

Before: Remainder theorem · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.