Question

factorise the l^2-(m-n)^2
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Answer 1

Answer:

हिंदी में खोजें

factorise एल 2- (mn 2

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How to solve your question

Your question is

2−1(−)2

l^{2}-1(m-n)^{2}l2−1(m−n)2

Simplify

1

Expand the square

2−1(−)2

l^{2}-1(m-n)^{2}l2−1(m−n)2

2−1(−)(−)

l^{2}-1(m-n)(m-n)l2−1(m−n)(m−n)

2

Distribute

2−1(−)(−)

l^{2}-1{\color{#c92786}{(m-n)(m-n)}}l2−1(m−n)(m−n)

2−1((−)−(−))

l^{2}-1({\color{#c92786}{m(m-n)-n(m-n)}})l2−1(m(m−n)−n(m−n))

3

Distribute

2−1((−)−(−))

l^{2}-1({\color{#c92786}{m(m-n)}}-n(m-n))l2−1(m(m−n)−n(m−n))

2−1(2−−(−))

l^{2}-1({\color{#c92786}{m^{2}-mn}}-n(m-n))l2−1(m2−mn−n(m−n))

4

Distribute

2−1(2−−(−))

l^{2}-1(m^{2}-mn{\color{#c92786}{-n(m-n)}})l2−1(m2−mn−n(m−n))

2−1(2−−+2)

l^{2}-1(m^{2}-mn{\color{#c92786}{-mn+n^{2}}})l2−1(m2−mn−mn+n2)

5

Combine like terms

2−1(2−−+2)

l^{2}-1(m^{2}{\color{#c92786}{-mn}}{\color{#c92786}{-mn}}+n^{2})l2−1(m2−mn−mn+n2)

2−1(2−2+2)

l^{2}-1(m^{2}{\color{#c92786}{-2mn}}+n^{2})l2−1(m2−2mn+n2)

6

Distribute

2−1(2−2+2)

l^{2}{\color{#c92786}{-1(m^{2}-2mn+n^{2})}}l2−1(m2−2mn+n2)

2−12+2−12

l^{2}{\color{#c92786}{-1m^{2}+2mn-1n^{2}}}l2−1m2+2mn−1n2

Solution

2−12+2−12

Step-by-step explanation:

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