Can anyone help me with the 14th question?Also,can anyone solve it on the paper?
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Answered by
1
Hope u like my process
=====================
Formula to be used ::
=-=-=-=-=-=-=-=-=-=-=-=-
=> (a - b) ³ = a³ - b³ - 3ab(a-b)
=> (a - b) ² = a² + b² - 2ab
__________________
If f(x) = x³ - 2x² + 7x + 5
Then ,
f(x-2)

Swipe left (<<==) to get the full line...
_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_
Hope this is ur required answer
Proud to help you
=====================
Formula to be used ::
=-=-=-=-=-=-=-=-=-=-=-=-
=> (a - b) ³ = a³ - b³ - 3ab(a-b)
=> (a - b) ² = a² + b² - 2ab
__________________
If f(x) = x³ - 2x² + 7x + 5
Then ,
f(x-2)
Swipe left (<<==) to get the full line...
_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_
Hope this is ur required answer
Proud to help you
rakeshmohata:
welcome.. Mera फर्ज़ था
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