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3.1 Algebra

Syllabus
9709–2028–2029
Topic
3.1
Level
A2

Modulus equations require separate sign cases and original-domain checks

For |f(x)|=g(x), first require g(x)≥0, then solve f(x)=g(x) or f(x)=−g(x). For inequalities, split according to whether the comparison side is non-negative.

Nested or variable-right-side moduli may need several intervals. Keep endpoints and test each candidate in the unsquared original equation.

|x−2|=x gives x≥0 and branches x−2=x (impossible) or 2−x=x, giving x=1.

Squaring both sides without checking the sign of the right side can add invalid solutions.

Polynomial division and factorisation reduce higher-degree equations safely

Use a known root a to divide P(x) by x−a, obtaining a lower-degree quotient. Then factor or solve the quotient and retain only roots satisfying the original polynomial.

Use the remainder theorem to test candidates, include missing powers as zero coefficients, and verify by multiplication. Repeated division reveals multiplicity.

If P(−1)=0 for a quartic, divide by x+1; if the quotient also vanishes at −1, the factor is repeated.

A factor found by trial does not guarantee the remaining quotient is quadratic; degree bookkeeping must be preserved.

Factor and remainder theorems expose roots of a higher-degree polynomial

For polynomial P, P(a)=0 exactly when x−a is a factor. Dividing by x−a leaves remainder P(a), so a quick substitution tests a candidate root.

After finding one factor, divide it out and solve the quotient. Repeated roots can be detected by testing the quotient or by checking derivatives when appropriate.

If P(−2)=0, x+2 divides P; a cubic then becomes a quadratic quotient that can be solved exactly.

The sign is tied to the factor: root a gives x−a, while root −a gives x+a.

Partial fractions rewrite a rational expression into integrable pieces

Decompose a proper rational function into simpler fractions whose denominators are the linear or irreducible factors of the original denominator. Repeated factors require repeated terms.

Factor the denominator first, write the correct unknown numerators, clear denominators and equate coefficients. If the fraction is improper, divide before decomposing.

1/[x(x+1)] = 1/x − 1/(x+1), making its integral ln|x|−ln|x+1|+C.

A repeated factor such as (x−1)² needs A/(x−1)+B/(x−1)²; one term cannot represent both powers.

Binomial coefficients identify a chosen power without expanding everything

In (a+b)^n the term containing a^{n−r}b^r is C(n,r)a^{n−r}b^r. The general term is useful for coefficients, approximations and selected powers.

Match the requested power before simplifying, include signs in b, and check whether n is a non-negative integer for a finite expansion.

The coefficient of x³ in (2−x)^5 is C(5,3)·2²·(−1)³=−40.

The row of Pascal’s triangle supplies only C(n,r); powers and signs still belong to the term.

Objective notes

5 learning objectives
ConceptA-Level CAIE Mathematics A2