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Algebra

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  Cambridge International AS & A Level Mathematics 9709 — Pure Mathematics 3

Algebra

The big idea: Most of Paper 3 algebra is about rewriting an expression into a more useful shape: a modulus into cases, a polynomial into factors, a fraction into partial fractions, a power into a series.

Summary — What This Chapter Covers
  • |x| is the distance of x from 0. The graph of y = |ax + b| is a V with its vertex on the x-axis at x = −b/a.
  • Key relations: |a| = |b| ⇔ a2 = b2, and |x − a| < b ⇔ a − b < x < a + b.
  • Polynomial division (degree up to 4, by a linear or quadratic divisor) gives a quotient and a remainder.
  • Factor theorem: (ax + b) is a factor of p(x) ⇔ p(−b/a) = 0. Remainder theorem: the remainder on dividing by (ax + b) is p(−b/a).
  • Partial fractions for denominators (ax + b)(cx + d)(ex + f), (ax + b)(cx + d)2 and (ax + b)(cx2 + d).
  • (1 + x)n for rational n is an infinite series, valid only for |x| < 1.
1. Modulus Functions and Inequalities
Modulus

Squaring, cases and sketches

|x| = x when x ≥ 0 and |x| = −x when x < 0. To sketch y = |ax + b|, draw y = ax + b and reflect the part below the x-axis in the x-axis, giving a V shape.

When both sides are moduli, square: both sides are non-negative so squaring keeps the equation or inequality equivalent. When only one side is a modulus, either split into cases (keeping each case's condition) or sketch both graphs and find where they cross.

Relations to use |a| = |b| ⇔ a2 = b2     |x − a| < b ⇔ a − b < x < a + b     |x − a| > b ⇔ x < a − b or x > a + b
Worked Example

Solve |3x − 2| = |2x + 7|.

Practice Question 1

Solve the inequality 2x + 5 < |x + 1|.

Practice Question 2

Solve |x − 3| < 2.

2. Polynomial Division, Factor and Remainder Theorems
Polynomials

Quotient, remainder and factors

Dividing p(x) by a divisor d(x) gives p(x) = d(x) × q(x) + r(x), where the remainder r(x) has lower degree than d(x). A linear divisor leaves a constant remainder; a quadratic divisor leaves a remainder of the form Ax + B. Use long division, or compare coefficients in the identity.

The factor and remainder theorems let you find remainders without dividing, and find unknown coefficients by forming simultaneous equations. The divisor may have a coefficient of x that is not 1, such as (2x + 1): substitute x = −½.

Theorems Remainder on dividing p(x) by (ax + b) = p(−b/a)     (ax + b) is a factor ⇔ p(−b/a) = 0
Worked Example

Find the quotient and remainder when x4 + 2x3 − x + 3 is divided by x2 + 1.

Practice Question

The polynomial p(x) = 2x3 + ax2 − 11x + b, where a and b are constants. (x − 2) is a factor of p(x), and when p(x) is divided by (x + 1) the remainder is 6. Find a and b, and factorise p(x) completely.

3. Partial Fractions
Partial fractions

Choosing the right form

A proper rational function (numerator degree lower than denominator degree) can be split into simpler fractions. Choose the form from the denominator, then multiply through and find the constants by substituting convenient values of x and comparing coefficients.

Forms 2 2 2 2
Worked Example

2

Practice Question 1

2

Practice Question 2

2 2

Memorise
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  • (1 + x)n when n is not a positive integer
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