1. **Problem:** Factorize the following expressions: i) $2x^2 - 16xy + 24y^2$ ii) $18x^2 - 128y^2$ iii) $2a^2y - 16a^3y^2 + 12ay^3$ iv) $64a^3 + 125b^3$ v) $a^3 + 5a = 6$ 2. **Formula and rules:** - Factorization involves expressing a polynomial as a product of simpler polynomials. - Common methods include taking out the greatest common factor (GCF), difference of squares, sum/difference of cubes, and factoring quadratics. 3. **Step-by-step factorization:** i) $2x^2 - 16xy + 24y^2$ - Take out GCF $2$: $$2x^2 - 16xy + 24y^2 = 2(x^2 - 8xy + 12y^2)$$ - Factor quadratic inside parentheses: Find two numbers that multiply to $12$ and add to $-8$: $-6$ and $-2$ $$2(x - 6y)(x - 2y)$$ ii) $18x^2 - 128y^2$ - Take out GCF $2$: $$2(9x^2 - 64y^2)$$ - Recognize difference of squares: $$9x^2 - 64y^2 = (3x)^2 - (8y)^2 = (3x - 8y)(3x + 8y)$$ - So: $$2(3x - 8y)(3x + 8y)$$ iii) $2a^2y - 16a^3y^2 + 12ay^3$ - Take out GCF $2ay$: $$2ay(a - 8a^2y + 6y^2)$$ - Inside parentheses, factor $a$ from first two terms: $$a - 8a^2y + 6y^2 = a(1 - 8ay) + 6y^2$$ - No simple factorization further; expression is: $$2ay(a - 8a^2y + 6y^2)$$ iv) $64a^3 + 125b^3$ - Recognize sum of cubes: $$64a^3 = (4a)^3, \, 125b^3 = (5b)^3$$ - Sum of cubes formula: $$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$ - So: $$(4a + 5b)(16a^2 - 20ab + 25b^2)$$ v) $a^3 + 5a = 6$ - Bring all terms to one side: $$a^3 + 5a - 6 = 0$$ - Try to factor by grouping or find roots: Test $a=1$: $$1 + 5 - 6 = 0$$ So, $(a - 1)$ is a factor. - Divide polynomial by $(a - 1)$: $$a^3 + 5a - 6 = (a - 1)(a^2 + a + 6)$$ 4. **Final answers:** i) $$2(x - 6y)(x - 2y)$$ ii) $$2(3x - 8y)(3x + 8y)$$ iii) $$2ay(a - 8a^2y + 6y^2)$$ iv) $$(4a + 5b)(16a^2 - 20ab + 25b^2)$$ v) $$(a - 1)(a^2 + a + 6) = 0$$