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1. Factorise. (a) \( a^{2}-4 a b+4 b^{2}-16 x^{2} \) (b) \( c^{2}-a^{2}+2 a b-b^{2} \) (c) \( a^{2}+2 a b+b^{2}-9 x^{2} \) (d) \( a^{2}-2 x y-y^{2}-x^{2} \) e) \( 16 x^{4}-a^{2}+6 a b-9 b^{2} \) (f) \( 25 y^{4}-1-9 m^{2}+6 m \) (g) \( 9 r^{2}-4+12 a-9 a^{2} \) (h) \( 1-81 a^{2}-18 a b-b^{2} \) i) \( 20 a b+4 a^{2}+25 b^{2}-9 x^{4} \) j) \( 24 x y-3 z^{2}+3 y^{2}+48 x^{2} \)

Ask by Harmon Peters. in South Africa
Feb 08,2025

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Answer

(a) \( (a - 2b - 4x)(a - 2b + 4x) \) (b) \( (c - a + b)(c + a - b) \) (c) \( (a + b - 3x)(a + b + 3x) \) (d) \( (a - x - y)(a + x + y) \) (e) \( (4x² - a + 3b)(4x² + a - 3b) \) (f) \( (5y² - 3m + 1)(5y² + 3m - 1) \)

Solution

We will factorise each expression step‐by‐step. ───────────────────────────── Example (a) Factorise   a² – 4ab + 4b² – 16x² Step 1. Notice that the first three terms form a perfect square:   a² – 4ab + 4b² = (a – 2b)² Step 2. Then your expression becomes a difference of two squares:   (a – 2b)² – (4x)² Step 3. Factor using the difference‐of‐squares formula:   X² – Y² = (X – Y)(X + Y) Thus,   (a – 2b – 4x)(a – 2b + 4x) ───────────────────────────── Example (b) Factorise   c² – a² + 2ab – b² Step 1. Group the last three terms:   a² – 2ab + b² = (a – b)² So rewrite the expression as:   c² – (a – b)² Step 2. Again a difference of two squares:   = (c – (a – b)) (c + (a – b)) which simplifies to   = (c – a + b)(c + a – b) ───────────────────────────── Example (c) Factorise   a² + 2ab + b² – 9x² Step 1. Recognize a² + 2ab + b² = (a + b)², so:   = (a + b)² – (3x)² Step 2. Factor the difference of two squares:   = (a + b – 3x)(a + b + 3x) ───────────────────────────── Example (d) Factorise   a² – 2xy – y² – x² Step 1. Notice that the last three terms can be regrouped:   x² + 2xy + y² = (x + y)² Thus, write:   a² – (x + y)² Step 2. Factor as a difference of squares:   = (a – (x + y))(a + (x + y))   = (a – x – y)(a + x + y) ───────────────────────────── Example (e) Factorise   16x⁴ – a² + 6ab – 9b² Step 1. Group the last three terms. Note that:   a² – 6ab + 9b² = (a – 3b)² So write:   16x⁴ – (a – 3b)² Step 2. Recognise 16x⁴ = (4x²)², then use difference of two squares:   = (4x² – (a – 3b)) (4x² + (a – 3b))   = (4x² – a + 3b)(4x² + a – 3b) ───────────────────────────── Example (f) Factorise   25y⁴ – 1 – 9m² + 6m Step 1. Rearrange the terms to group two pairs:   (25y⁴ – 1) – (9m² – 6m) Step 2. Notice that 25y⁴ – 1 is a difference of two squares,   25y⁴ – 1 = (5y²)² – 1² = (5y² – 1)(5y² + 1)   However, the second group does not factor nicely by itself. A better idea is to rewrite the last three terms as a complete square:   9m² – 6m + 1 = (3m – 1)²  so that   –(9m² – 6m + 1) = –(3m – 1)² Step 3. Now write the whole expression as:   25y⁴ – (3m – 1)² Step 4. Then factor as a difference of squares:   = (5y² – (3m – 1)) (5y² + (3m – 1))   = (5y² – 3m + 1)(5y² + 3m – 1) ───────────────────────────── Example (g) Factorise   9r² – 4 + 12a – 9a² Step 1. Rearrange the terms to group those in a:   9r² – 9a² + 12a – 4 Step 2. Notice that 9a² – 12a + 4 = (3a – 2)². Thus write:   9r² – (3a – 2)² Step 3. Factor as a difference of two squares:   = (3r – (3a – 2)) (3r + (3a – 2))   = (3r – 3a + 2)(3r + 3a – 2) ───────────────────────────── Example (h) Factorise   1 – 81a² – 18ab – b² Step 1. Group the last three terms. Notice that:   81a² + 18ab + b² = (9a + b)² Step 2. Then the expression becomes:   1 – (9a + b)² Step 3. Factor as a difference of two squares:   = (1 – (9a + b))(1 + (9a + b))   = (1 – 9a – b)(1 + 9a + b) ───────────────────────────── Example (i) Factorise   20ab + 4a² + 25b² – 9x⁴ Step 1. Rearrange the terms so that the a–b part is together:   4a² + 20ab + 25b² = (2a + 5b)² Step 2. Then the expression becomes:   (2a + 5b)² – 9x⁴   and 9x⁴ = (3x²)² Step 3. Factor by difference of squares:   = (2a + 5b – 3x²)(2a + 5b + 3x²) ───────────────────────────── Example (j) Factorise   24xy – 3z² + 3y² + 48x² Step 1. Group the terms in x and y:   48x² + 24xy + 3y² Notice that this is a perfect square since   48x² + 24xy + 3y² = 3(16x² + 8xy + y²) = 3(4x + y)² Step 2. Now the whole expression becomes:   3(4x + y)² – 3z² Step 3. Factor out the common factor 3 and use the difference of squares:   = 3[(4x + y)² – z²]   = 3[(4x + y) – z][(4x + y) + z]   = 3(4x + y – z)(4x + y + z) ───────────────────────────── Summary of Answers (a) (a – 2b – 4x)(a – 2b + 4x) (b) (c – a + b)(c + a – b) (c) (a + b – 3x)(a + b + 3x) (d) (a – x – y)(a + x + y) (e) (4x² – a + 3b)(4x² + a – 3b) (f) (5y² – 3m + 1)(5y² + 3m – 1)

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Bonus Knowledge

To tackle these factorisation problems, let's take a look at two of the more intricate concepts involved in solving them! First up, understanding how to group terms effectively is key to simplifying complex polynomials. If you observe patterns such as perfect squares or the difference of squares, you can regroup your terms into manageable chunks. For example, recognizing \(a^2 - 4b^2\) as a difference of squares can lead to quicker solutions! Next, don't shy away from checking your results through expansion! Once you've created potential factors, it's always a good idea to multiply them back together to ensure they yield the original expression. This step not only solidifies your understanding but reinforces your problem-solving skills. Plus, who doesn’t love validating their hard work with a little bit of multiplication?

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