Algebraic expressions and fractions are the bedrock of Leaving Certificate Higher Level Mathematics. Success here requires mastering systematic methods for polynomial division, factorisation, binomial expansion, and fraction manipulation. Securing full marks depends on maintaining a clean layout, naming your theorems, and showing every intermediate step.
Polynomial Long Division and Theorems
Polynomial long division follows the same layout as numerical long division: divide, multiply, subtract, bring down. Before starting, write both the dividend and divisor in descending powers of x. If any term is missing, insert it with a zero coefficient (e.g., 0x²). A remainder of zero proves the divisor is a factor. The Factor Theorem states that for a polynomial f(x), (x − c) is a factor if and only if f(c) = 0, which gives you an easy way to verify roots before doing the heavy lifting of long division. For a divisor with a leading coefficient like (ax − b), you evaluate f(b/a) instead. The Remainder Theorem works on the same logic: dividing f(x) by (x − c) leaves a remainder of f(c) without running through the whole division calculation. Neither theorem is printed in your Formulae and Tables booklet, so you need to know both statements by heart.
Essential Factorisation Identities
Order of attack: HCF first → count the terms. If you have two terms, check for a difference of two squares, a² − b² = (a − b)(a + b). If they are cubed, use a³ ± b³ = (a ± b)(a² ∓ ab + b²) and remember the SOAP sign rule: Same sign, Opposite sign, Always Positive. For three terms, use the guide-number method: take 6x² + x − 15. The guide number is 6 × (−15) = −90, and the two factors multiplying to −90 while adding to +1 are +10 and −9. Splitting that middle term gives 6x² + 10x − 9x − 15, which factors by grouping into 2x(3x + 5) − 3(3x + 5) = (3x + 5)(2x − 3). For four terms, use grouping: xy − 2y + 3x − 6 = y(x − 2) + 3(x − 2) = (x − 2)(y + 3).
Binomial Theorem and Expansions
The Binomial Theorem expands any power (x + y)^n into a sum of terms using combinatorial coefficients: (x + y)^n = Σ_{r=0}^{n} C(n, r) x^{n−r} y^r. The exam rarely asks for the entire expansion; instead, you will use the general term formula, T_{r+1} = C(n, r) x^{n−r} y^r, to pull out a specific term. In every term, the powers of x and y must add to n. When finding the independent term, set the total exponent of x to zero and solve for r. When evaluating C(n, r), write the symbol form before the calculator step. Be careful: the whole coefficient and sign of the term (like -3/x²) must be raised to the power r.
Algebraic Fractions and Rational Expressions
Algebraic fractions obey exactly the same rules as numerical fractions. Always factorise denominators first to find the lowest common denominator (LCD). When solving equations, multiply every term on both sides by the LCD and reject any excluded values that make a denominator zero. When simplifying, only cancel factors that divide the whole numerator and whole denominator. Never cancel across a + or − sign. Keep an eye out for swapped denominators like (b − a) when another fraction has (a − b). Pull out a minus sign using b − a = −(a − b), which instantly gives you a matching common denominator.
Key terms
- Factor Theorem
- The algebraic result stating that (x − c) is a factor of a polynomial f(x) precisely when f(c) = 0.
- Remainder Theorem
- The rule showing that dividing any polynomial f(x) by a linear divisor (x − c) leaves a remainder equal to f(c).
- Quotient
- The polynomial obtained when the dividend is divided by the divisor, satisfying dividend = divisor × quotient + remainder.
- Equating coefficients
- Two polynomials are equal for all x if and only if the coefficients of matching powers of x are equal.
- Binomial Theorem
- The algebraic expansion formula expressing (x + y)^n as a sum of terms using combinatorial coefficients C(n, r).
- Excluded values
- Values of x that make a denominator zero, rendering the expression undefined.
Check yourself
Factorise 27x³ − 8.
(3x − 2)(9x² + 6x + 4).
If (x − 3) is a factor of x³ + kx² − 4x − 3, find k.
f(3) = 0 ⟹ 27 + 9k − 12 − 3 = 0 ⟹ 9k = −12 ⟹ k = −4/3.
Find the coefficient of x³ in (1 + 2x)⁵.
T₄ = C(5, 3)(1)²(2x)³ = 10 × 1 × 8x³ = 80x³. Coefficient = 80.
