Algebra is a fundamental area of the Leaving Certificate Mathematics syllabus, appearing in both Paper 1 and Paper 2. This topic covers algebraic expressions, equations, inequalities, complex Numbers, matrices, and sequences.

Expansion is the process of removing brackets by multiplying each term inside by the expression Outside.

Example (OL/HL): Expand $(2x + 3)(x - 5)$.

$$(2x + 3)(x - 5) = 2x^2 - 10x + 3x - 15 = 2x^2 - 7x - 15$$

Example (HL): Expand and simplify $(x + 2)^3$.

$$(x + 2)^3 = (x + 2)(x + 2)^2 = (x + 2)(x^2 + 4x + 4) = x^3 + 4x^2 + 4x + 2x^2 + 8x + 8 = x^3 + 6x^2 + 12x + 8$$

Factorisation is the reverse process. Common techniques include:

1. Common factor: $6x^2 + 9x = 3x(2x + 3)$
2. Quadratic trinomial: $x^2 + 5x + 6 = (x + 2)(x + 3)$
3. Difference of two squares: $x^2 - 16 = (x - 4)(x + 4)$

The identity $a^2 - b^2 = (a - b)(a + b)$ extends to factorisation:

$$4x^2 - 25y^2 = (2x - 5y)(2x + 5y)$$

Proof. $(a-b)(a+b) = a^2 + ab - ab - b^2 = a^2 - b^2$.

This identity also leads to the sum and difference of two cubes:

$$A^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ $$A^3 + b^3 = (a + b)(a^2 - ab + b^2)$$

Verification: $(a - b)(a^2 + ab + b^2) = a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3 = a^3 - b^3$.

Example (HL): Factorise $x^4 - 16$ completely.

$$X^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$$

Example (HL): Factorise $x^3 + 8$.

$$X^3 + 8 = (x + 2)(x^2 - 2x + 4)$$$$A^2 + 2ab + b^2 = (a + b)^2$$ $$A^2 - 2ab + b^2 = (a - b)^2$$

Example (HL): Factorise $9x^2 - 30x + 25$.

$$9x^2 - 30x + 25 = (3x - 5)^2$$

Solve for $x$: $3(2x - 1) = 4(x + 3) - 5$.

$$6x - 3 = 4x + 12 - 5$$ $$6x - 4x = 7 + 3$$ $$2x = 10 \implies x = 5$$

Example (HL): Solve $\frac{2x + 1}{3} - \frac{x - 2}{4} = \frac{1}{6}$.

Multiply through by 12 (the LCM of 3, 4, 6):

$$4(2x + 1) - 3(x - 2) = 2$$ $$8x + 4 - 3x + 6 = 2$$ $$5x + 10 = 2 \implies 5x = -8 \implies x = -\frac{8}{5}$$

Quadratic equations take the form $ax^2 + bx + c = 0$. Methods of solution include:

X^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0 \implies x = 2 \mathrm{ or x = 3

For $ax^2 + bx + c = 0$:

$$X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Derivation. Complete the square for $ax^2 + bx + c = 0$:

$a\!\left(x^2 + \frac{b}{a}x\right) = -c$

$a\!\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} = -c$

$a\!\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a}$

$x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}$

Example: Solve $2x^2 + 3x - 5 = 0$.

Here $a = 2$, $b = 3$, $c = -5$.

$$X = \frac{-3 \pm \sqrt{9 + 40}}{4} = \frac{-3 \pm 7}{4}$$ X = 1 \mathrm{ or x = -\frac{5}{2}

Write $ax^2 + bx + c$ in the form $a(x - h)^2 + k$.

Example: Express $x^2 + 6x + 2$ in completed square form.

$$X^2 + 6x + 2 = (x^2 + 6x + 9) - 9 + 2 = (x + 3)^2 - 7$$

The discriminant $\Delta = b^2 - 4ac$ determines the nature of the roots:

Condition	Roots
$\Delta > 0$	Two distinct real roots
$\Delta = 0$	One repeated real root
$\Delta < 0$	No real roots (two complex roots)

Example (HL): Find the range of $k$ for which $x^2 + 4x + k = 0$ has real roots.

$$\Delta = 16 - 4k \geq 0 \implies k \leq 4$$

Example (HL): Find the range of $k$ for which $kx^2 + 4x + k = 0$ has real roots.

$$\Delta = 16 - 4k^2 \geq 0 \implies k^2 \leq 4 \implies -2 \leq k \leq 2$$

Example (HL): Find the value of $k$ for which $x^2 + 2kx + 9 = 0$ has equal roots.

$$\Delta = 4k^2 - 36 = 0 \implies k^2 = 9 \implies k = \pm 3$$

Example (OL): Solve:

$$\begin{cases} 2x + y = 7 \\ X - y = 2 \end{cases}$$

Adding: $3x = 9 \implies x = 3$. Substituting: $y = 1$.

Example (HL): Solve:

$$\begin{cases} X + y = 5 \\ X^2 + y^2 = 13 \end{cases}$$

From the first equation $y = 5 - x$. Substituting:

$$X^2 + (5 - x)^2 = 13$$ $$X^2 + 25 - 10x + x^2 = 13$$ $$2x^2 - 10x + 12 = 0 \implies x^2 - 5x + 6 = 0$$ (x - 2)(x - 3) = 0 \implies x = 2 \mathrm{ or x = 3

Solutions: $(2, 3)$ and $(3, 2)$.

When multiplying or dividing both sides by a negative number, reverse the inequality sign.

Example: Solve $3 - 2x > 7$.

$$-2x > 4 \implies x < -2$$

Example: Solve $x^2 - 3x - 4 < 0$.

Factorise: $(x - 4)(x + 1) < 0$.

The product is negative when one factor is positive and the other negative. Since the parabola opens Upward:

$$-1 < x < 4$$

:::caution Always check the inequality sign carefully. For $x^2 - 3x - 4 \gt 0$The solution would Be $x \lt -1$ or $x \gt 4$ (outside the roots).

Example (HL): Solve $\frac{x - 1}{x + 2} \le 0$.

Critical values: $x = 1$ (numerator zero) and $x = -2$ (denominator zero).

Sign chart:

Interval	$x < -2$	$-2 < x < 1$	$x > 1$
$x - 1$	Negative	Negative	Positive
$x + 2$	Negative	Positive	Positive
Quotient	Positive	Negative	Non-negative

Solution: $x \in (-2, 1]$.

$|x| \lt a \iff -a \lt x \lt a$

$|x| > a \iff x \lt -a$ or $x > a$

Example: Solve $|2x - 3| \lt 5$.

$-5 \lt 2x - 3 \lt 5$

$-2 \lt 2x \lt 8$

$-1 \lt x \lt 4$

A complex number is of the form $z = a + bi$ where $a, b \in \mathbb{R}$ and $i^2 = -1$.

• $a$ is the real part ($\operatorname{Re}(z)$)
• $b$ is the imaginary part ($\operatorname{Im}(z)$)

Addition: $(a + bi) + (c + di) = (a + c) + (b + d)i$

Multiplication: $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$

Complex conjugate: $\bar{z} = a - bi$

$$Z \cdot \bar{z} = a^2 + b^2$$

Division: $\dfrac{a + bi}{c + di} = \dfrac{(a + bi)(c - di)}{(c + di)(c - di)} = \dfrac{(ac + bd) + (bc - ad)i}{c^2 + d^2}$

The modulus (or absolute value) of $z = a + bi$:

$$|z| = \sqrt{a^2 + b^2}$$

The argument $\arg(z)$ is the angle $\theta$ measured from the positive real axis:

$$\theta = \arctan\left(\frac{b}{a}\right), \quad a > 0$$

For $a < 0$Add $\pi$ to get the correct quadrant.

Quadrant check for the argument:

| Quadrant | $a$ | $b$ | $\arg(z)$ | | -------- | ----- | ----- | --------------------- | --- | --- | | I | $> 0$ | $> 0$ | $\arctan(b/a)$ | | II | $< 0$ | $> 0$ | $\pi - \arctan(b/ | a | )$ | | III | $< 0$ | $< 0$ | $-\pi + \arctan(b/a)$ | | IV | $> 0$ | $< 0$ | $-\arctan(b/a)$ |

$$Z = r(\cos\theta + i\sin\theta) = re^{i\theta}$$

Where $r = |z|$ and $\theta = \arg(z)$.

Example (HL): Express $z = -1 + i\sqrt{3}$ in polar form.

$r = \sqrt{1 + 3} = 2$. Since $a = -1 < 0$ and $b = \sqrt{3} > 0$The point is in the second Quadrant.

$$\theta = \pi - \arctan\!\left(\frac{\sqrt{3}}{1}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$ $$Z = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = 2e^{2\pi i/3}$$

For $z = r(\cos\theta + i\sin\theta)$ and $n \in \mathbb{Z}$:

$$Z^n = r^n(\cos n\theta + i\sin n\theta)$$

Proof by induction for positive integers. Base case $n = 1$: trivial. Inductive step: assume True for $n = k$. Then $z^{k+1} = z^k \cdot z = r^k(\cos k\theta + i\sin k\theta) \cdot r(\cos\theta + i\sin\theta)$. Expanding using addition formulae gives $r^{k+1}(\cos(k+1)\theta + i\sin(k+1)\theta)$.

Example: Express $(\sqrt{3} + i)^4$ in the form $a + bi$.

First find modulus and argument: $r = 2$, $\theta = \frac{\pi}{6}$.

$$(\sqrt{3} + i)^4 = 2^4\left(\cos\frac{4\pi}{6} + i\sin\frac{4\pi}{6}\right) = 16\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = 16\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -8 + 8\sqrt{3}\,i$$

Example (HL): Express $z = 1 - i\sqrt{3}$ in polar form and hence find $z^5$.

$r = \sqrt{1 + 3} = 2$. Since $a = 1 > 0$ and $b = -\sqrt{3} < 0$The point is in the fourth Quadrant.

$$\theta = -\frac{\pi}{3}, \quad z = 2e^{-\pi i/3}$$ $$Z^5 = 2^5 e^{-5\pi i/3} = 32\left(\cos\frac{-5\pi}{3} + i\sin\frac{-5\pi}{3}\right) = 32\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 16 - 16\sqrt{3}\,i$$

The $n$Th roots of unity are the solutions to $z^n = 1$.

$$Z_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}, \quad k = 0, 1, 2, \ldots, n-1$$

These lie on the unit circle in the complex plane, equally spaced at angles of $\frac{2\pi}{n}$.

Properties:

• The sum of all $n$Th roots of unity is $0$.
• The product of all $n$Th roots of unity is $(-1)^{n-1}$.
• The $n$Th roots of any complex number $w = re^{i\theta}$ are $\sqrt[n]{r}\, e^{i(\theta + 2k\pi)/n}$ for $k = 0, 1, \ldots, n-1$.

Example: Find the cube roots of unity.

For $z^3 = 1$: $z_k = \cos\frac{2k\pi}{3} + i\sin\frac{2k\pi}{3}$, $k = 0, 1, 2$.

$$Z_0 = 1, \quad z_1 = -\frac{1}{2} + i\frac{\sqrt{3}}{2}, \quad z_2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$$

Note: $1 + z_1 + z_2 = 0$.

Example (HL): Find all complex numbers $z$ such that $z^4 = 16$.

The four fourth roots of $16 = 16e^{0i}$ are:

$$Z_k = 2\,e^{2k\pi i/4}, \quad k = 0, 1, 2, 3$$ $$Z_0 = 2, \quad z_1 = 2i, \quad z_2 = -2, \quad z_3 = -2i$$

A matrix $A$ of order $m \times n$ has $m$ rows and $n$ columns.

Addition: Add corresponding elements (matrices must be the same order).

Scalar multiplication: Multiply every element by the scalar.

Matrix multiplication: If $A$ is $m \times p$ and $B$ is $p \times n$Then $AB$ is $m \times n$.

The $(i, j)$ entry of $AB$ is:

$$(AB)_{ij} = \sum_{k=1}^{p} a_{ik}b_{kj}$$

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

$$\det(A) = ad - bc$$

If $\det(A) \neq 0$:

$$A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$

For $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & k \end{pmatrix}$:

$$\det(A) = a(ek - fh) - b(dk - fg) + c(dh - eg)$$

For $AX = B$ where $A$ is invertible:

$$X = A^{-1}B$$

Example: Solve:

$$\begin{cases} 2x + y = 5 \\ X - y = 1 \end{cases}$$ $$\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$$ $$\det(A) = -2 - 1 = -3$$ $$A^{-1} = \frac{1}{-3}\begin{pmatrix} -1 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{pmatrix}$$ $$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{pmatrix}\begin{pmatrix} 5 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$

1. Base case: Show the statement holds for $n = 1$ (or the smallest relevant value).
2. Inductive hypothesis: Assume the statement holds for $n = k$.
3. Inductive step: Show that if it holds for $n = k$It also holds for $n = k + 1$.
4. Conclusion: By the principle of mathematical induction, the statement holds for all $n \geq 1$.

Why induction works. The base case anchors the chain at $n = 1$. The inductive step shows that If any link in the chain holds, the next one does too. Together, they prove that every link holds.

Example: Prove by induction that $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$.

Base case ($n = 1$): LHS $= 1$RHS $= \frac{1 \cdot 2 \cdot 3}{6} = 1$. True.

Inductive hypothesis: Assume $\sum_{r=1}^{k} r^2 = \frac{k(k+1)(2k+1)}{6}$.

Inductive step:

$$\sum_{r=1}^{k+1} r^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$$ $$= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6}$$ $$= \frac{(k+1)[k(2k+1) + 6(k+1)]}{6}$$ $$= \frac{(k+1)(2k^2 + 7k + 6)}{6}$$ $$= \frac{(k+1)(k+2)(2k+3)}{6}$$ $$= \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$

This matches the formula with $n = k + 1$. By induction, the result holds for all $n \ge 1$.

Example: Prove that $\sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4}$.

Base case ($n = 1$): LHS $= 1$RHS $= \frac{1 \cdot 4}{4} = 1$. True.

Inductive hypothesis: Assume $\sum_{r=1}^{k} r^3 = \frac{k^2(k+1)^2}{4}$.

Inductive step:

$$\sum_{r=1}^{k+1} r^3 = \frac{k^2(k+1)^2}{4} + (k+1)^3$$ $$= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}$$ $$= \frac{(k+1)^2[k^2 + 4(k+1)]}{4}$$ $$= \frac{(k+1)^2(k^2 + 4k + 4)}{4}$$ $$= \frac{(k+1)^2(k+2)^2}{4}$$ $$= \frac{(k+1)^2((k+1)+1)^2}{4}$$

By induction, the result holds for all $n \ge 1$.

To divide $P(x)$ by $(x - a)$Use either long division or synthetic division. The result gives:

$$P(x) = (x - a)Q(x) + R$$

Where $Q(x)$ is the quotient and $R$ is the remainder. By the Remainder Theorem, $R = P(a)$.

Factor Theorem: $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$.

Example: Factorise $x^3 - 3x + 2$.

Try $P(1) = 1 - 3 + 2 = 0$So $(x - 1)$ is a factor.

Dividing: $x^3 - 3x + 2 = (x - 1)(x^2 + x - 2) = (x - 1)(x + 2)(x - 1) = (x - 1)^2(x + 2)$.

Example (HL): When $P(x) = x^3 + 2x^2 - 5x - 6$ is divided by $(x - 1)$The remainder is $P(1) = 1 + 2 - 5 - 6 = -8$.

Example: Solve $(x - 1)^2(x + 2) \gt 0$.

The critical values are $x = -2$ and $x = 1$ (double root).

Sign chart:

Interval	$x < -2$	$-2 < x < 1$	$x > 1$
$(x+2)$	Negative	Positive	Positive
$(x-1)^2$	Positive	Positive	Positive
Product	Negative	Positive	Positive

Solution: $x < -2$ or $x > 1$I.e., $x \in (-\infty, -2) \cup (1, \infty)$.

Note that $x = -1$ is not a solution (the product equals zero, not positive). And $x = 1$ is not a Solution despite being a root, because the factor is squared.

See the examples integrated throughout the sections above.

1. Sign errors in factorisation and expansion — always double-check by expanding back.
2. Forgetting to reverse the inequality when multiplying/dividing by a negative number.
3. Confusing the discriminant conditions for real vs. Complex roots. $\Delta < 0$ means two complex conjugate roots, not “no roots.”
4. De Moivre’s theorem requires the argument to be in radians.
5. Matrix multiplication is not commutative: $AB \neq BA$ .
6. Induction base case — always state and verify it explicitly. A proof without a base case is like a chain with no anchor.
7. Domain restrictions on logarithms — $\log_a(x)$ is only defined for $x > 0$.
8. Forgetting absolute values in the quadratic formula when $\Delta < 0$: $x = \frac{-b \pm i\sqrt{|\Delta|}}{2a}$.
9. Confusing the argument quadrant. For $z = -1 + i$ (second quadrant), $\arg(z) = \frac{3\pi}{4}$Not $\arctan(-1)$.

1. Expand $(3x - 2)(2x + 5)$.
2. Factorise $x^2 - 7x + 12$.
3. Solve $4x - 7 = 2x + 9$.
4. Solve $x^2 - 5x + 6 = 0$ by factorisation.
5. Solve $2x^2 + x - 3 = 0$ using the quadratic formula.

1. Prove by induction that $\sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4}$.
2. Express $z = 1 - i\sqrt{3}$ in polar form and hence find $z^5$.
3. Find the modulus and argument of $\frac{1 + i}{1 - i}$.
4. Find the values of $k$ for which $kx^2 + 4x + k = 0$ has equal roots.
5. Given $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & -1 \\ 2 & 3 \end{pmatrix}$Find $AB - BA$.
6. Find all complex numbers $z$ such that $z^4 = 16$.
7. Solve the inequality $x^2 - 2x - 15 \gt 0$.
8. Factorise $x^4 - 1$ completely.
9. Find the remainder when $P(x) = 2x^3 - 3x^2 + 5x - 7$ is divided by $(x + 2)$.
10. Solve $|3x - 1| \le 8$.
11. Find the modulus and argument of $z = \frac{3 + 4i}{1 - 2i}$.
12. Prove by induction that $3^n \ge 2^n + n$ for all $n \ge 1$.
13. Express $\frac{2x + 1}{(x+1)(x-2)}$ in partial fractions.
14. Solve $z^3 = -27$ and plot all solutions on an Argand diagram.
15. Find the quadratic equation whose roots are $2 + \sqrt{3}$ and $2 - \sqrt{3}$.

16. Given $z = 2 + 3i$ and $w = 1 - 4i$Find $z\bar{w}$ and $|z/w|$.
17. Solve the simultaneous equations $x + iy + iz = 0$ and $x - 2y + iz = 1 + i$ for real $x$ and $y$.
18. Prove by induction that $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1}$.
19. Find the matrix $A$ such that $A\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ and $A\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 7 \\ 7 \end{pmatrix}$.
20. Express $\frac{3x^2 - x + 2}{(x-1)(x^2 + 1)}$ in partial fractions.
21. Find all complex numbers $z$ satisfying $|z - 2i| = |z + 2|$ and interpret geometrically.
22. Prove that if a quadratic equation with rational coefficients has one irrational root $a + b\sqrt{c}$ (where $b \neq 0$), then $a - b\sqrt{c}$ is also a root.

This topic covers the mathematical techniques and concepts related to algebra, including key theorems, methods, and problem-solving approaches.

Key concepts include:

• quadratic equations and the discriminant
• simultaneous equations
• polynomial division and the factor theorem
• partial fractions
• binomial expansion

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

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