Pregunta

1.2. Factorise fully: 1.2.1. \( a^{2}-2 a-a x+2 x \) 1.2.2. \( -x y-(y-x) b+b^{2} \) 1.2.3. \( 6 p+40-p^{2} \) 1.2.4. \( a^{2}-2 a b+b^{2}-100 c^{2} \) 1.2.5. \( 2 x^{2}-x-6 \)

Ask by Horton Smith. in South Africa
Feb 25,2025

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1. (a – 2)(a – x) 2. (b + x)(b – y) 3. (10 – p)(p + 4) 4. (a – b – 10c)(a – b + 10c) 5. (x – 2)(2x + 3)

Solución

We will factorise each expression step‐by‐step. ──────────────────────────── 1.2.1. Factorise a² – 2a – ax + 2x Step 1. Group the terms:   a² – 2a – ax + 2x = (a² – 2a) + (–ax + 2x) Step 2. Factor each group:   a² – 2a = a(a – 2)   –ax + 2x = –x(a – 2)    (factor out –x) Step 3. Factor out the common factor (a – 2):   a(a – 2) – x(a – 2) = (a – 2)(a – x) Answer: (a – 2)(a – x) ──────────────────────────── 1.2.2. Factorise –xy – (y – x)b + b² Step 1. Expand the middle term:   –xy – (y – x)b + b² = –xy – by + bx + b² Step 2. Rearrange and group:   = b² + bx – by – xy   Group as: [b² + bx] + [– by – xy] Step 3. Factor each group:   b² + bx = b(b + x)   –by – xy = –y(b + x) Step 4. Factor out the common factor (b + x):   b(b + x) – y(b + x) = (b + x)(b – y) Answer: (b + x)(b – y) ──────────────────────────── 1.2.3. Factorise 6p + 40 – p² Step 1. Write in standard form (highest power first):   –p² + 6p + 40 Step 2. Factor out –1:   = –(p² – 6p – 40) Step 3. Factor the quadratic inside the parentheses.   Find two numbers whose product is (–40) and whose sum is –6.   Numbers –10 and 4 work because (–10) × 4 = –40 and (–10) + 4 = –6.   Express:    p² – 6p – 40 = p² – 10p + 4p – 40   Group:    = p(p – 10) + 4(p – 10)   = (p – 10)(p + 4) Step 4. Bring back the overall factor –1:   –(p – 10)(p + 4) can also be written as (10 – p)(p + 4) Answer: (10 – p)(p + 4) ──────────────────────────── 1.2.4. Factorise a² – 2ab + b² – 100c² Step 1. Recognise the perfect square:   a² – 2ab + b² = (a – b)² So the expression becomes:   (a – b)² – 100c² Step 2. Recognise a difference of two squares:   100c² = (10c)² Then:   (a – b)² – (10c)² = (a – b – 10c)(a – b + 10c) Answer: (a – b – 10c)(a – b + 10c) ──────────────────────────── 1.2.5. Factorise 2x² – x – 6 Step 1. Multiply the coefficient of x² by the constant term:   2 × (–6) = –12 Step 2. Find two numbers that multiply to –12 and add to –1 (the coefficient of x):   The numbers –4 and 3 work because –4 × 3 = –12 and –4 + 3 = –1. Step 3. Rewrite and group:   2x² – x – 6 = 2x² – 4x + 3x – 6   Group terms:    = (2x² – 4x) + (3x – 6)   Factorize each group:    2x² – 4x = 2x(x – 2)    3x – 6 = 3(x – 2) Step 4. Factor out the common factor (x – 2):   = (x – 2)(2x + 3) Answer: (x – 2)(2x + 3) ──────────────────────────── Final Answers: 1. (a – 2)(a – x) 2. (b + x)(b – y) 3. (10 – p)(p + 4) 4. (a – b – 10c)(a – b + 10c) 5. (x – 2)(2x + 3)

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To factorise the given expressions fully: 1.2.1. \( a^{2}-2a-ax+2x \) Rearranging gives us \( a^{2} - 2a - ax + 2x = a(a-2-x) + 2x \). We can rewrite it as \( a(a-2-x) + 2(x)\). Thus, the expression becomes \( (a-2)(a-x+2) \). 1.2.2. \( -xy-(y-x)b+b^{2} \) Rearranging gives us \( -xy - by + bx + b^{2} \) or \( -y(x-b) + b(x+b) \). Factoring out common terms leads to \( -(x-b)(y+b) \). 1.2.3. \( 6p + 40 - p^{2} \) Rearranging gives us \( -p^{2} + 6p + 40 = -(p^{2} - 6p - 40) \). Factoring the quadractic gives us \( -(p-10)(p+4) \). Thus, the expression becomes \( -(p-10)(p+4) \). 1.2.4. \( a^{2}-2ab+b^{2}-100c^{2} \) We'll notice the first part is a perfect square: \( (a-b)^{2} - (10c)^{2} \), which is a difference of squares. It can be factored as \( ((a-b)-10c)((a-b)+10c) \). 1.2.5. \( 2x^{2}-x-6 \) We will factor the quadratic. Start with \( (2x+3)(x-2) \) as it satisfies \( 2x^2 - x - 6 \). Thus, the expression is \( 2(x + 3)(x - 2) \).

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