Question

2.
Factorise: 2 (ab-cd)+a^2+b^2-c^2-d^2
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Answer 1

Answer:

(a+b+c+d)(a+b-c-d)

Step-by-step explanation:

2(ab-cd)+a²+b²-c²-d²

= 2ab-2cd+a²+b²-c²-d²

= (a²+b²+2ab)-c²-d²-2cd

= (a²+b²+2ab)-(c²+d²+2cd)

= (a+b)²-(c+d)²

= [(a+b)+(c+d)][(a+b)-(c+d)]

=(a+b+c+d)(a+b-c-d)

Therefore,

2(ab-cd)+a²+b²-c²-d²

= (a+b+c+d)(a+b-c-d)

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Answer 2

(a + b + c + d) (a + b - c - d)

Step-by-step explanation:

Given,

2(ab - cd) + a² + b² - c² - d²

Multiplying by 2:

2ab - 2cd + a² + b² - c² - d²

Arranging the given expression:

a² + b² + 2ab - c² - d² - 2cd

(a² + b² + 2ab) - (c² + d² + 2cd)

Applying the formula: a² + b² + 2ab = (a + b)²

(a + b)² - (c + d)²

Applying the formula: a² - b² = (a + b) - (a - b)

[(a + b) + (c + d)] [(a + b) - (c + d)]

(a + b + c + d) (a + b - c - d)

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