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Question
Factorize using appropriate identity:
t2−2t+1
t2−2t+1
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Solution
The given expression is t2−2t+1.We know that, a2−2ab+b2=(a−b)2Here, a=t and b=1Therefore, t2−2t+1=(t)2−2(t)(1)+(1)2⇒(t−1)2⇒(t−1)(t−1)
Hence, t2−2t+1=(t−1)(t−1).
The given expression is t2−2t+1.
We know that, a2−2ab+b2=(a−b)2
Here, a=t and b=1
Therefore, t2−2t+1=(t)2−2(t)(1)+(1)2
⇒(t−1)2
⇒(t−1)(t−1)
Hence, t2−2t+1=(t−1)(t−1).
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