Emma Nie
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Solomon paper f
Q9a)
Q3b)
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RDKGames
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(Original post by Emma Nie)
Solomon paper f
Q9a)
Q3b)
So urgent* that you do not have time to attach these questions and your attempt at them?
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Emma Nie
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Q3b) not understanding why you need to squre 8 and times the and from a)
Q9a)I do not have a clue of why the MS do that shouldn't it be like factory therom?
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Philip-flop
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(Original post by RDKGames)
So urgent* that you do not have time to attach these questions and your attempt at them?
Even more than that! OP didn't even have time to reach for the 'u' key when spelling the word urgent!
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Emma Nie
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(Original post by Emma Nie)
Solomon paper f
Q9a)
Q3b)
Attached files
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Apachai Hopachai
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Question 9a)

The 'factory' theorem states that if we have a polynomial, say   p(x) then if  p(a) = 0 where a is just some number then  (x-a) is a factor of  p(x) . A factor means  p(x) can be written in the form  (x-a)f(x)

This theorem cannot be used to form the polynomial. However, if you recall a geometric series is a series which takes the form  \underbrace{ \{a,ar,ar^2,ar^3,..,.ar^{n-1}\}}_{n\text{ terms}}

 

 For each term, you just multiply by  r . Therefore,  \frac{ar^2}{ar} = \frac{ar}{a}

Try applying this to your question
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Emma Nie
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(Original post by Apachai Hopachai)
Question 9a)

The 'factory' theorem states that if we have a polynomial, say   p(x) then if  p(a) = 0 where a is just some number then  (x-a) is a factor of  p(x) . A factor means  p(x) can be written in the form  (x-a)f(x)

This theorem cannot be used to form the polynomial. However, if you recall a geometric series is a series which takes the form  \underbrace{ \{a,ar,ar^2,ar^3,..,.ar^{n-1}\}}_{n\text{ terms}}

 

 For each term, you just multiply by  r . Therefore,  \frac{ar^2}{ar} = \frac{ar}{a}

Try applying this to your question
Get it now. How about Q3?I am confused with how to calculate the area
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