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secondary 3 | A Maths
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how to solve part ii? thank u!
Date Posted: 4 years ago
Views: 207
(a + b)⁴
= a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
In this case, your 2x = a and 1/5x = b
Notice that the the x terms cancel out each other if you multiply a by b (2x by 1/5x)
So if you square that , a²b² also has all the x terms cancelled out.
So the term independent of x is 6a²b²
= a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
In this case, your 2x = a and 1/5x = b
Notice that the the x terms cancel out each other if you multiply a by b (2x by 1/5x)
So if you square that , a²b² also has all the x terms cancelled out.
So the term independent of x is 6a²b²
thank you!
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done
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Since this question is asked as a ‘hence’ question, I replaced a and b by the respective terms and write all the five terms out manually.
Date Posted:
4 years ago
thank you!
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