Simplifying Complex Algebraic Expressions
Simplifying Complex Algebraic Expressions
Grade 11 Mathematics • G.C.E. O/L Sri Lanka Local Syllabus
Learning Objectives
- Combine positive, negative, and fractional indices in multi-step algebraic simplifications.
- Convert composite numerical bases (\(4, 8, 9, 16, 25, 27, 32\)) into prime factors before applying index laws.
- Convert radical (root) expressions into fractional index form to simplify rational terms.
Essential Prime Base Conversions
In O/L exam questions, composite bases must be rewritten as powers of prime numbers (\(2, 3, 5\)) so the bases match:
| Base 2 | Base 3 | Base 5 |
|---|---|---|
| \(4 = 2^2\) | \(9 = 3^2\) | \(25 = 5^2\) |
| \(8 = 2^3\) | \(27 = 3^3\) | \(125 = 5^3\) |
| \(16 = 2^4\) | \(81 = 3^4\) | \(625 = 5^4\) |
| \(32 = 2^5\) | \(243 = 3^5\) | — |
Step-by-Step Worked Examples
- Convert the square root \(\sqrt{x^3}\) into fractional index form \((x^3)^{\frac{1}{2}} = x^{\frac{3}{2}}\): \[\frac{x^{\frac{3}{2}} \times x^{-\frac{1}{2}}}{x^{\frac{1}{3}}}\]
- Apply the product law in the numerator (\(x^{\frac{3}{2} + (-\frac{1}{2})} = x^{\frac{2}{2}} = x^1\)): \[\frac{x^1}{x^{\frac{1}{3}}}\]
- Apply the quotient law: \[x^{1 – \frac{1}{3}} = x^{\frac{2}{3}}\]
- Express all composite bases (\(4\) and \(8\)) in prime base \(2\): \[\frac{(2^2)^{n+1} \times 2^{n-2}}{(2^3)^n} = \frac{2^{2n+2} \times 2^{n-2}}{2^{3n}}\]
- Combine the numerator exponents using the product law: \[\frac{2^{(2n+2) + (n-2)}}{2^{3n}} = \frac{2^{3n}}{2^{3n}}\]
- Apply the quotient law and zero index rule: \[2^{3n – 3n} = 2^0 = 1\]
Practice Drill
Show Step-by-Step Answer
Step 1: Convert \(9\) and \(27\) to base \(3\) (\(9 = 3^2\), \(27 = 3^3\)):
\(\frac{(3^2)^{n+1} \times 3^{n-1}}{(3^3)^n} = \frac{3^{2n+2} \times 3^{n-1}}{3^{3n}}\)
Step 2: Apply product law in numerator and quotient law:
\(\frac{3^{3n+1}}{3^{3n}} = 3^{(3n+1) – 3n} = 3^1 = \mathbf{3}\)
Show Step-by-Step Answer
Step 1: Rewrite the cube root as a fractional power \(\frac{1}{3}\) and write \(8 = 2^3\):
\((2^3 a^6 b^{-3})^{\frac{1}{3}}\)
Step 2: Multiply each inner exponent by \(\frac{1}{3}\):
\(2^{3 \times \frac{1}{3}} \cdot a^{6 \times \frac{1}{3}} \cdot b^{-3 \times \frac{1}{3}} = 2^1 a^2 b^{-1} = \mathbf{\frac{2a^2}{b}}\)
Show Step-by-Step Answer
Step 1: Rewrite \(\sqrt[4]{x^3}\) as \(x^{\frac{3}{4}}\):
\(\frac{x^{\frac{1}{2}} \times x^{\frac{3}{4}}}{x^{-\frac{1}{4}}}\)
Step 2: Combine all exponents using product and quotient laws:
\(x^{\frac{1}{2} + \frac{3}{4} – (-\frac{1}{4})} = x^{\frac{2}{4} + \frac{3}{4} + \frac{1}{4}} = x^{\frac{6}{4}} = \mathbf{x^{\frac{3}{2}}}\)