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Good evening Candice! Here are my workings for the first two parts.
Date Posted:
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.
Thank you so much
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 -
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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Part ii, but these two substitutions take some observations.
Date Posted:
4 years ago