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secondary 3 | A Maths
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Ok um I sorta got the ans but got p = 9 and p = 1 and idk how to explain y p = 1 is rej cuz the ans key never like explain
Date Posted: 2 years ago
Views: 369
You have to see which value of p does not work with “x + 3”.
One of them will result in a remainder which is non-zero when divided by (x + 3).
One of them will result in a remainder which is non-zero when divided by (x + 3).
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done
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clear
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Use my presentation or use the factor theorem to prove that p cannot be 1
Date Posted:
2 years ago
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