Algebra - Expressions & Factorising

Leaving Cert Higher Level Mathematics revision notes with key terms and self-check questions.

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Algebra at Leaving Certificate Higher Level centres on manipulating polynomial and rational expressions with speed and precision. On Paper 1, questions rarely announce which technique to choose. You must look at the structure of an expression first—spotting highest common factors, quadratic forms, difference of two squares, or cube patterns before putting pen to paper. Master the factor theorem for dividing higher-degree polynomials and the binomial general term formula for finding individual coefficients without expanding whole brackets.

Algebraic Structure and Polynomial Operations

A polynomial is an expression where every power of the variable is a non-negative whole number (0,1,2,…0, 1, 2, \dots). The degree of the polynomial is its highest exponent. When you add or subtract polynomials, group strictly by like terms: matching letters with identical powers.

When multiplying terms, coefficients multiply while indices add according to index rules (xa×xb=xa+bx^a \times x^b = x^{a+b}). For long division of polynomials, arrange both dividend and divisor in descending powers of xx. If a power is missing, insert it with a coefficient of zero (such as +0x2+ 0x^2) so your columns align properly. The division identity states:

Dividend=(Divisor×Quotient)+Remainder\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}

If the remainder turns out to be zero, the divisor divides evenly and is confirmed as a factor of the original polynomial.

Factorising Strategies and Special Patterns

Always scan for a highest common factor (HCF) before trying any advanced method. Pulling out an HCF simplifies the numbers and immediately reveals the underlying structure.

For four terms, group them into pairs that share common factors. For quadratics ax2+bx+cax^2 + bx + c, use the guide number method: multiply a×ca \times c, find two factors of this product that sum to bb, split the middle term, and finish by grouping.

Recognising standard algebraic identities saves vital time on Paper 1:

  • Difference of two squares: a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)
  • Sum of two cubes: a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2)
  • Difference of two cubes: a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Notice the sign pattern for cubes: the first bracket shares the sign of the problem, the middle term of the quadratic bracket takes the opposite sign, and the end term is always positive.

Rational Algebraic Expressions

Rational expressions are fractions that contain algebraic polynomials in their numerator, denominator, or both. They follow the exact same arithmetic rules as numerical fractions.

To add or subtract, find the lowest common denominator (LCD), rewrite each fraction with this denominator, and combine numerators over the common base. Never expand the denominator unless the question explicitly asks you to write the answer in a specific expanded form. Keeping the denominator factorised lets you see if any common brackets can cancel at the end.

When multiplying or dividing, factorise every numerator and denominator completely before doing anything else. Divide out common factors between top and bottom. To divide by a fraction, multiply by its reciprocal (invert the second fraction). Watch out for expressions like (2−x)(2 - x) and (x−2)(x - 2); they are negatives of each other, meaning (2−x)=−1(x−2)(2 - x) = -1(x - 2).

The Factor Theorem and Polynomial Roots

The Factor Theorem connects polynomial division directly to roots: a linear expression (x−c)(x - c) is a factor of a polynomial P(x)P(x) if and only if P(c)=0P(c) = 0.

If you need to factorise a cubic polynomial P(x)P(x) from scratch, test integer factors of the constant term (such as ±1,±2,±3\pm 1, \pm 2, \pm 3) until you find a value cc where P(c)=0P(c) = 0. Once you find a root, (x−c)(x - c) is your first factor. Divide P(x)P(x) by (x−c)(x - c) using algebraic long division or by equating coefficients. The resulting quotient will be a quadratic expression, which you can then factorise into two linear brackets or solve using the quadratic formula.

If (ax−b)(ax - b) is a factor, substituting x=bax = \frac{b}{a} into the polynomial will yield zero. When an exam question states that (x−c)(x - c) leaves a remainder RR, set P(c)=RP(c) = R to solve for missing coefficients.

The Binomial Theorem

For any positive whole number nn, the binomial expansion of (x+y)n(x + y)^n contains exactly n+1n + 1 terms:

(x+y)n=∑r=0n(nr)xn−ryr(x + y)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^r

As you move left to right, powers of xx decrease from nn to 00, while powers of yy increase from 00 to nn. The general term formula from the Formulae and Tables booklet gives the (r+1)th(r + 1)^{\text{th}} term:

Tr+1=(nr)xn−ryrT_{r+1} = \binom{n}{r} x^{n-r} y^r

To find a specific term without expanding the whole bracket, select the correct rr. For example, the 4th term uses r=3r = 3. If the binomial contains a minus sign, such as (2x−3)n(2x - 3)^n, treat yy as (−3)(-3) so negative signs appear with their correct powers. To locate a term independent of xx, write out Tr+1T_{r+1}, gather all powers of xx, set that combined index to zero, and solve for rr.

Key terms

Polynomial
An algebraic expression composed of terms with non-negative whole-number exponents.
Degree
The value of the highest exponent of the variable in a polynomial.
Factor Theorem
A theorem stating that a polynomial P(x) has a factor (x - c) if and only if P(c) = 0.
Quotient
The result obtained from dividing one polynomial expression by another.
Rational Expression
An algebraic fraction where both numerator and denominator are polynomial expressions.
Binomial Coefficient
The numerical factor in a binomial term, written as nCr or (n choose r).
Term Independent of x
A constant term in an algebraic expansion where the exponent of x equals zero.

Check yourself

  1. If (x - 2) is a factor of f(x) = 2x³ - 3x² + kx - 6, find the value of k.

    Apply the Factor Theorem: f(2) = 0. So 2(2)³ - 3(2)² + k(2) - 6 = 0 => 16 - 12 + 2k - 6 = 0 => 2k - 2 = 0 => k = 1.

  2. Factorise the difference of cubes 8x³ - 27 fully.

    Write as (2x)³ - (3)³. Applying a³ - b³ = (a - b)(a² + ab + b²) gives (2x - 3)(4x² + 6x + 9).

  3. What is the coefficient of x³ in the expansion of (1 + 2x)⁵?

    Use T_{r+1} with r = 3: ⁵C₃ (1)² (2x)³ = 10 · 1 · 8x³ = 80x³. The coefficient is 80.

  4. Simplify the rational expression (x² - 9)/(2x² + 5x - 3).

    Factorise top and bottom: (x - 3)(x + 3) / [(2x - 1)(x + 3)]. Cancel the common factor (x + 3) to get (x - 3)/(2x - 1).

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