Question
EXERCISE 6 (a) Factorise: \( \begin{array}{lll}\text { (1) } 6 x^{3}+12 x & \text { (2) } 6 x^{3}+4 x^{2} & \text { (3) } 5 x^{3}+5 x \\ \text { (4) } 12 x^{3}-18 x^{2} & \text { (5) } & 3 x^{2}-9 y+12 x y\end{array} \) \( \begin{array}{ll}\text { (7) } 16 m^{4} n^{8}-8 m^{3} n^{7}+36 m^{2} n^{3} & \\ \text { (7) } & \\ \text { (6 } b^{2}-64 a b\end{array} \)
Ask by Zimmerman Carrillo. in South Africa
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
Here are the factorizations:
1. \(6x^{3} + 12x = 6x(x^{2} + 2)\)
2. \(6x^{3} + 4x^{2} = 2x^{2}(3x + 2)\)
3. \(5x^{3} + 5x = 5x(x^{2} + 1)\)
4. \(12x^{3} - 18x^{2} = 6x^{2}(2x - 3)\)
5. \(3x^{2} - 9y + 12xy = 3(x^{2} + 4xy - 3y)\)
6. \(b^{2} - 64ab = b(b - 64a)\)
7. \(16m^{4}n^{8} - 8m^{3}n^{7} + 36m^{2}n^{3} = 4m^{2}n^{3}(4m^{2}n^{5} - 2mn^{4} + 9)\)
Solution
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The Deep Dive
Sure! Here are the factorizations for each expression: 1. \( 6x^3 + 12x = 6x(x^2 + 2) \) 2. \( 6x^3 + 4x^2 = 2x^2(3x + 2) \) 3. \( 5x^3 + 5x = 5x(x^2 + 1) \) 4. \( 12x^3 - 18x^2 = 6x^2(2x - 3) \) 5. \( 3x^2 - 9y + 12xy = 3(x^2 + 4xy - 3y) \) (This last expression could be further factored if you'd like to factor the quadratic.) 6. \( b^2 - 64ab = b(b - 64a) \) 7. \( 16m^4n^8 - 8m^3n^7 + 36m^2n^3 = 4m^2n^3(4mn^5 - 2mn^4 + 9) \) Let me know if you need any more help!