Question

$factorise \\ \\ 1)(l + m) {}^{2} - 4lm$
if your answer is right then I will give you brainliest answer

please give answer step by step​

Answer 1

Solution :-

Method -1:-

Given that (l+m)² - 4lm

=> (l²+2lm+m²)- 4lm

Since, (a+b)² = a²+2ab+b²

=> l²+m²+(2lm-4lm)

=> l²+m²-2lm

=> (l-m)²

Since, (a-b)² = a²-2ab+b²

=> (l-m)(l-m)

Therefore,

(l+m)² - 4lm = (l-m)(l-m)

Method-2:-

Given that (l+m)² - 4lm

We know that

(a-b)² = (a+b)² - 4ab

Therefore, (l+m)² - 4lm = (l-m)²

=> (l-m)(l-m)

Therefore,

(l+m)² - 4lm = (l-m)(l-m)

Answer 2

FACTORISE (l+m)² − 4lm

SOLUTION :-

First way to solve this :-

We have, (l + m) ^ 2 - 4lm

= (l² + 2lm + m²) - 4lm

(Using (a + b)² = a² +b² +2ab)

= l² + (2lm − 4lm) + m²

(Combining the like terms)

= l² =2lm + m²

= (l)² - 2(l)(m) + (m)²

= (l -m)² (Using a² +b² - 2ab = (a-b) ² )

second way to solve this :-

(l + m)² - 4lm = l² + m² + 2 l m - 4lm

{Using p² + 2pq + q² = (p+q)²}

= l² + m² - 2lm

= l²- 2lm + m²

=(l - m)² = (l - m)(l - m)

{Using p²- 2pq + q² = (p - q)²}