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\begin{tabular}{l} (a) Simplify \( \frac{2}{y^{0}} \) \\ (b) Simplify fully \( \left(16 a^{4}\right)^{\frac{3}{4}} \) \\ 5 \\ Factorise fully \( 18 c^{3} d^{2}-21 c^{2} \) \\ (Total for Question 4 is 3 marks) \\ \hline\end{tabular}

Ask by Howell Osborne. in the United Arab Emirates
Jan 22,2025

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Tutor-Verified Answer

Answer

- **(a)** \( 2 \) - **(b)** \( 8a^{3} \) - **Factorised Form:** \( 3c^{2} (6c d^{2} - 7) \)

Solution

Sure, let's solve each part step by step. --- ### (a) Simplify \( \frac{2}{y^{0}} \) Any non-zero number raised to the power of 0 is 1. Therefore: \[ \frac{2}{y^{0}} = \frac{2}{1} = 2 \] **Answer:** \( 2 \) --- ### (b) Simplify fully \( \left(16 a^{4}\right)^{\frac{3}{4}} \) Apply the exponentiation rule \((x^{m})^{n} = x^{m \times n}\): \[ \left(16 a^{4}\right)^{\frac{3}{4}} = 16^{\frac{3}{4}} \times \left(a^{4}\right)^{\frac{3}{4}} \] Simplify each part separately: 1. **Simplify \(16^{\frac{3}{4}}\):** \[ 16 = 2^{4} \Rightarrow 16^{\frac{3}{4}} = \left(2^{4}\right)^{\frac{3}{4}} = 2^{3} = 8 \] 2. **Simplify \(\left(a^{4}\right)^{\frac{3}{4}}\):** \[ \left(a^{4}\right)^{\frac{3}{4}} = a^{3} \] Combine both results: \[ \left(16 a^{4}\right)^{\frac{3}{4}} = 8a^{3} \] **Answer:** \( 8a^{3} \) --- ### (5) Factorise fully \( 18 c^{3} d^{2} - 21 c^{2} \) 1. **Find the greatest common factor (GCF):** - Coefficients: GCF of 18 and 21 is 3. - Variables: Both terms have \(c^{2}\). 2. **Factor out the GCF:** \[ 18 c^{3} d^{2} - 21 c^{2} = 3c^{2} \left(6c d^{2} - 7\right) \] **Answer:** \( 3c^{2} (6c d^{2} - 7) \) --- **Summary of Answers:** - **(a)** \( 2 \) - **(b)** \( 8a^{3} \) - **Factorised Form:** \( 3c^{2} (6c d^{2} - 7) \)

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To simplify \( \frac{2}{y^{0}} \), remember that any non-zero number raised to the power of 0 equals 1. Thus, \( y^0 = 1 \), so \( \frac{2}{y^{0}} = \frac{2}{1} = 2 \). For \( \left(16 a^{4}\right)^{\frac{3}{4}} \), first, rewrite it using the property of exponents: \( (x^m)^n = x^{m \cdot n} \). Therefore, \( \left(16^{\frac{3}{4}}\right) \cdot \left(a^{4 \cdot \frac{3}{4}}\right) = 16^{\frac{3}{4}} \cdot a^3 \). Since \( 16 = 2^4 \), it becomes \( (2^4)^{\frac{3}{4}} = 2^{3} = 8 \), so the result is \( 8a^{3} \). Next, to factorise \( 18 c^{3} d^{2}-21 c^{2} \), first notice that both terms share common factors. The greatest common factor (GCF) is 3 \( c^{2} \), so factor that out: \( 3 c^{2} (6 cd - 7) \). Voilà! You have the fully factored expression.

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