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Additional Mathematics Synthesis Guide 6: Polynomials, Factor and Remainder Theorem, Partial Fractions and Binomial Expansion

BUKIT TIMAH TUTOR · ADDITIONAL MATHEMATICS · SYNTHESIS GUIDE 6

Polynomial algebra is a study of structure: how expressions factor, divide, decompose and expand while preserving the same mathematical object.

Factor theorem, remainder theorem, partial fractions and binomial expansion can look like four separate chapters. They are connected by one deeper idea: rewrite an algebraic object into a form that makes a required property easier to see or use. A polynomial can be evaluated through its remainder. A factor can be detected without long division. A rational expression can be decomposed into simpler pieces. A power of a binomial can be expanded systematically without repeated multiplication.

Polynomial → evaluate → factor → divide → decompose → expand → verify.

1. Polynomials are organised by powers

A polynomial in x is built from terms anxn + … + a1x + a0 with non-negative integer powers. The degree is the highest power with a non-zero coefficient. Structure matters because division, factorisation and roots are connected.

If P(x) = (x − a)Q(x), then x = a is a root of P(x), because P(a)=0. This is the algebraic bridge between factors and roots.

2. The remainder theorem compresses polynomial division

When P(x) is divided by x − a, the remainder is P(a). This means one substitution can answer a division question without performing the whole division.

For P(x)=2x3−5x+7, the remainder on division by x−2 is

P(2)=2(8)−10+7=13.

3. The factor theorem is the zero-remainder case

If P(a)=0, then x−a is a factor of P(x). The factor theorem is therefore not a separate magical rule. It is the remainder theorem with remainder zero.

This connection matters because it helps students remember what the theorem is doing rather than memorising a sentence mechanically.

4. Worked Example 1 — test a factor

Show that x−2 is a factor of P(x)=x3−3x2−4x+12.

P(2)=8−12−8+12=0. Therefore x−2 is a factor.

Dividing gives P(x)=(x−2)(x2−x−6)=(x−2)(x−3)(x+2).

5. Unknown coefficients turn theorem questions into simultaneous conditions

Suppose P(x)=x3+ax2+bx+6 and x−1 and x+2 are factors. Then P(1)=0 and P(−2)=0. These two factor conditions create two equations for a and b.

P(1)=1+a+b+6=0, so a+b=−7.

P(−2)=−8+4a−2b+6=0, so 2a−b=1.

Solving gives a=−2 and b=−5.

6. Repeated roots create repeated factors

If x=a is a repeated root of a polynomial, then (x−a)2 is a factor. In more advanced work, repeated roots can also be detected by the derivative condition P(a)=0 and P′(a)=0. This is an important bridge between polynomial algebra and calculus.

7. Partial fractions reverse a rational combination

A rational expression such as

(5x+1)/[(x−1)(x+2)]

can be rewritten in the simpler form

A/(x−1)+B/(x+2).

Combining the right-hand side gives

[A(x+2)+B(x−1)]/[(x−1)(x+2)].

Therefore the numerators must agree identically.

8. Worked Example 2 — decompose a rational expression

Decompose (5x+1)/[(x−1)(x+2)].

Let

(5x+1)/[(x−1)(x+2)] = A/(x−1)+B/(x+2).

Then 5x+1=A(x+2)+B(x−1).

Set x=1: 6=3A, so A=2.

Set x=−2: −9=−3B, so B=3.

Hence 2/(x−1)+3/(x+2).

9. Repeated factors change the partial-fraction template

If the denominator contains (x−a)2, the decomposition must include both

A/(x−a)+B/(x−a)2.

Leaving out one power creates an incomplete representation and makes coefficient matching impossible.

10. Irreducible quadratic factors require a linear numerator

For a factor such as x2+1 that does not split into real linear factors, the corresponding numerator takes the form Ax+B. This is another structural rule: the numerator degree must be less than the denominator-factor degree.

11. Improper fractions must be divided first

If the numerator degree is at least the denominator degree, perform polynomial division first. Partial fractions is designed for the proper rational remainder. Skipping this step often creates an impossible coefficient system.

12. Binomial expansion is controlled repeated multiplication

For a positive integer n,

(a+b)n = Σ C(n,r)an−rbr

for r from 0 to n. The coefficients come from combinations because each term counts how many ways b can be selected from the n factors.

The general term is

Tr+1 = C(n,r)an−rbr.

13. Worked Example 3 — expand and identify a term

Expand the first four terms of (2+x)5.

Using coefficients 1,5,10,10,…:

(2+x)5 = 32 + 80x + 80x2 + 40x3 + …

The coefficient structure is doing most of the work. Repeated manual multiplication would be slower and more error-prone.

14. Worked Example 4 — coefficient of a selected power

Find the coefficient of x3 in (1−2x)7.

Use r=3:

C(7,3)(1)4(−2x)3 = 35(−8)x3 = −280x3.

Coefficient = −280.

15. Independent term questions are exponent-balance questions

When x appears in both parts of a binomial, the general term can have an x-power depending on r. The term independent of x occurs when that total exponent is zero. This turns a coefficient question into a small equation in r.

16. Worked Example 5 — find a constant term

Find the term independent of x in (x2 + 2/x)6.

General term:

Tr+1=C(6,r)(x2)6−r(2x−1)r.

The exponent of x is 2(6−r)−r = 12−3r. Set 12−3r=0, so r=4.

Constant term = C(6,4)·24 = 15·16 = 240.

17. These topics connect through reversibility

Factorisation and expansion are inverse-style processes: one compresses products into an expression, the other exposes multiplicative structure. Partial fractions and combining rational expressions are also reverse processes. The factor and remainder theorems connect evaluation to division. Seeing these pairs reduces the number of isolated procedures a learner must hold.

Expansion ↔ factorisation; combination ↔ partial-fraction decomposition; division ↔ remainder evaluation.

18. Common failure patterns

  • Substituting the wrong sign into the factor theorem. Factor x+3 means test x=−3.
  • Claiming a factor because P(a) is small rather than exactly zero.
  • Forgetting repeated-factor partial-fraction terms.
  • Using a constant numerator over an irreducible quadratic.
  • Attempting partial fractions before polynomial division when the fraction is improper.
  • Confusing term number with r. Tr+1 uses r starting from zero.
  • Losing signs in (a−b)n.
  • Expanding everything when only one coefficient is needed.

19. A reliable structural routine

  1. Identify whether the problem asks about evaluation, factorisation, decomposition or expansion.
  2. For x−a divisors, test P(a).
  3. Translate factor conditions into equations for unknown coefficients.
  4. Before partial fractions, factor the denominator completely and check whether the rational expression is proper.
  5. Write the complete decomposition template before solving for constants.
  6. For binomial questions, write the general term before hunting for a coefficient or power.
  7. Use exponent conditions deliberately.
  8. Recombine, expand or substitute to verify.

20. Practice set

  1. Find the remainder when P(x)=x3+2x−1 is divided by x−2.
  2. Determine whether x+1 is a factor of x3−4x2+x+6.
  3. Factorise x3−6x2+11x−6 completely.
  4. If x−2 is a factor of x3+kx2−5x+6, find k.
  5. Decompose (3x+7)/[(x+1)(x+2)].
  6. Write the correct partial-fraction form for 1/[(x−1)2(x+3)].
  7. Write the correct numerator form over an irreducible factor x2+4.
  8. State why an improper rational expression should be divided before decomposition.
  9. Find the coefficient of x2 in (1+3x)5.
  10. Find the coefficient of x4 in (2−x)6.
  11. Find the constant term in (x+1/x)4.
  12. Find the term independent of x in (x2+3/x)6.
  13. Explain why factor theorem is a special case of remainder theorem.
  14. If P(4)=7, what is the remainder when P(x) is divided by x−4?
  15. If x+2 is a factor of P(x), what must P(−2) equal?
  16. Expand the first three terms of (3+x)4.
  17. Find the general term of (2+x)8.
  18. Why is partial-fraction decomposition useful before integration in later mathematics?
  19. How can recombining partial fractions verify the decomposition?
  20. How can substitution verify a polynomial factorisation?

Answers

  1. 11.
  2. P(−1)=0, so yes.
  3. (x−1)(x−2)(x−3).
  4. k=−1.
  5. 4/(x+1)−1/(x+2).
  6. A/(x−1)+B/(x−1)2+C/(x+3).
  7. Ax+B.
  8. Because the standard decomposition applies to a proper rational remainder whose numerator degree is lower than the denominator degree.
  9. 90.
  10. 60.
  11. 6.
  12. 1215.
  13. Because the remainder upon division by x−a is P(a); a factor occurs exactly when that remainder is zero.
  14. 7.
  15. 0.
  16. 81+108x+54x2+…
  17. C(8,r)28−rxr.
  18. Because simpler rational terms are easier to integrate separately.
  19. Put the terms over a common denominator and confirm the original numerator is recovered.
  20. Multiply the proposed factors or test their roots in the original polynomial.

21. What mastery looks like

Mastery means the student recognises what structure should be exposed. They know when evaluation is cheaper than division, when a zero remainder proves a factor, how denominator factorisation determines a partial-fraction template, and how the general binomial term can answer a targeted coefficient question without a full expansion.

The transfer test is to remove the chapter label. Present an expression and ask what representation would make the next task easiest. If the learner can choose between substitution, factorisation, division, decomposition and expansion for a reason, the algebra is becoming connected rather than procedural.

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