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secondary 4 | A Maths
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Candice
Candice

secondary 4 chevron_right A Maths chevron_right Singapore

Hello! Sorry for all the questions this late in the night

Date Posted: 4 years ago
Views: 501

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
Good evening Candice! Here are my workings for the first two parts.
Eric Nicholas K
Eric Nicholas K
4 years ago
Candice, there is no need to worry about posting the questions late! You can just post it as usual and someone will probably look at the questions if they are still awake or the next day.
Candice
Candice
4 years ago
Thank you so much
J
J
4 years ago
Interesting thing to note is that

2t³ - 5t² + t + 2 = (t - 1)(2t² - 3t - 2)
= (t - 1)(2t + 1)(t - 2)

While

2t³ + t² - 5t + 2 = (t - 1)(2t² + 3t - 2)
= (t - 1)(2t - 1)(t + 2)

The coefficients of t² and t are swapped and you can get from 1 expression to the other just by changing the sign of 3t from + to -
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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
Part ii, but these two substitutions take some observations.