Algebraic Expressions and Operations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your NECO exam.

Basic Definitions:

• Term: A single number, variable, or product of numbers and variables (e.g., $3x^2$, $-7y$, $5$)
• Expression: A combination of terms connected by $+$ or $-$ (e.g., $3x^2 - 7y + 5$)
• Equation: A statement that two expressions are equal (contains an $=$ sign)
• Identity: An equation that is true for all values of the variable (e.g., $(a+b)^2 = a^2 + 2ab + b^2$)

Laws of Indices:

Rule	Example
$a^m \times a^n = a^{m+n}$	$x^3 \times x^5 = x^8$
$a^m \div a^n = a^{m-n}$	$y^7 \div y^2 = y^5$
$(a^m)^n = a^{mn}$	$(z^3)^4 = z^{12}$
$a^0 = 1$ (for $a \neq 0$)	$5^0 = 1$
$a^{-n} = \frac{1}{a^n}$	$x^{-3} = \frac{1}{x^3}$
$a^{\frac{1}{n}} = \sqrt[n]{a}$	$25^{\frac{1}{2}} = 5$
$a^{\frac{m}{n}} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$	$8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4$

Expanding Brackets:

• Single bracket: $a(b + c) = ab + ac$
• Double bracket: $(a+b)(c+d) = ac + ad + bc + bd$
• Special products: $(a+b)^2 = a^2 + 2ab + b^2$; $(a-b)^2 = a^2 - 2ab + b^2$; $(a+b)(a-b) = a^2 - b^2$

⚡ NECO Tip: When simplifying algebraic fractions, factorise numerators and denominators first, then cancel common factors. For expanding $(2x+3)(5x-4)$: multiply each term of the first bracket by each term of the second: $10x^2 - 8x + 15x - 12 = 10x^2 + 7x - 12$.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for NECO Mathematics students with a few days to months.

Factorisation Techniques:

1. Common factor: $6x^2 + 9x = 3x(2x + 3)$
2. Difference of two squares: $a^2 - b^2 = (a+b)(a-b)$. Example: $x^2 - 16 = (x+4)(x-4)$
3. Trinomials: $x^2 + (p+q)x + pq = (x+p)(x+q)$ where $p$ and $q$ multiply to give constant term and add to give coefficient of $x$.
   • Example: $x^2 + 7x + 12 = (x+3)(x+4)$ (since $3+4=7$ and $3 \times 4=12$)
4. Quadratic trinomials: $acx^2 + (ad+bc)x + bd = (ax+b)(cx+d)$
   • Example: $6x^2 + 5x - 6 = (3x-2)(2x+3)$ (cross-multiply to verify: $3 \times 3 + (-2) \times 2 = 9 - 4 = 5$ ✓)

Algebraic Fractions:

$$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$$

To add or subtract fractions with different denominators: find the LCM of denominators.

Example: $\frac{3}{x+2} + \frac{5}{x-3}$ LHS: $\frac{3(x-3) + 5(x+2)}{(x+2)(x-3)} = \frac{3x-9+5x+10}{x^2-x-6} = \frac{8x+1}{x^2-x-6}$

Indices and Surds:

Simplify $\sqrt{50}$: $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$

Rationalising the denominator: $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$; $\frac{1}{3 + \sqrt{2}} = \frac{3 - \sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})} = 3 - \sqrt{2}$ (using difference of two squares)

Linear Equations:

Solve $3(x - 2) = 2(x + 5) + 7$: $3x - 6 = 2x + 10 + 7 \Rightarrow 3x - 6 = 2x + 17 \Rightarrow x = 23$

Quadratic Equations:

Using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Example: Solve $2x^2 - 5x - 3 = 0$: $a=2, b=-5, c=-3$; discriminant $= (-5)^2 - 4(2)(-3) = 25 + 24 = 49$ $x = \frac{5 \pm 7}{4} \Rightarrow x = 3$ or $x = -\frac{1}{2}$

⚡ NECO Common Mistakes:

• Not applying the distributive law correctly when expanding
• Forgetting to check whether factorisation is complete before solving
• In quadratic formula, getting the sign of $b$ wrong ($b$ is the coefficient as written in $ax^2 + bx + c$)
• Mixing up factorisation methods for different types of expressions

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for NECO and JAMB Mathematics preparation.

Partial Fractions:

Decompose $\dfrac{5x+1}{(x-1)(x+2)}$ into partial fractions: Let $\frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$ $\Rightarrow 5x+1 = A(x+2) + B(x-1)$ Set $x=1$: $6 = A(3) \Rightarrow A = 2$ Set $x=-2$: $-9 = B(-3) \Rightarrow B = 3$ $\Rightarrow \dfrac{5x+1}{(x-1)(x+2)} = \dfrac{2}{x-1} + \dfrac{3}{x+2}$

Completing the Square:

$x^2 + 6x + 5 = (x+3)^2 - 9 + 5 = (x+3)^2 - 4$

This form is useful for: (1) finding vertex of a parabola; (2) solving quadratic equations; (3) integrating rational functions.

Simultaneous Equations:

Solve: $2x + 3y = 7$ and $4x - y = 5$

By elimination: Multiply second equation by 3: $12x - 3y = 15$ Add to first equation: $14x = 22 \Rightarrow x = \frac{11}{7}$ Substitute back: $4(\frac{11}{7}) - y = 5 \Rightarrow y = \frac{44}{7} - 5 = \frac{9}{7}$

Determinants (2×2):

For $\begin{pmatrix}a & b \ c & d\end{pmatrix}$, determinant $= ad - bc$ If $ad - bc = 0$, the system has no unique solution (either no solution or infinitely many).

Indices Deep Dive:

$$a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$$ $$a^{-m} = \frac{1}{a^m}$$ $$a^p \cdot a^q = a^{p+q}$$

Example: Simplify $\dfrac{2^5 \times 2^{-3}}{2^4} = 2^{5-3-4} = 2^{-2} = \frac{1}{4}$

Logarithms and Indices:

Since $a^x = b \Leftrightarrow \log_a b = x$: $$\log_a (bc) = \log_a b + \log_a c$$ $$\log_a \frac{b}{c} = \log_a b - \log_a c$$ $$\log_a b^c = c \log_a b$$ $$\log_a a = 1, \quad \log_a 1 = 0$$

Change of Base: $$\log_a b = \frac{\log_c b}{\log_c a}$$

NECO/JAMB Patterns:

• NECO often asks: simplify expressions using laws of indices; factorise quadratic and cubic expressions; solve simultaneous linear equations; simplify algebraic fractions; rationalise denominators containing surds

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