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2002 Winter Diet, Paper 1, Q5 - HSN forum

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2002 Winter Diet, Paper 1, Q5


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#1 kr88

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Posted 18 April 2005 - 08:44 AM

does anyone have the working on their computer for this question? i have no problems with this kind of question but no matter how often i do the working i still get the wrong answer! v frustrating! i get c=-25 and d=18, when c actually =-19 and d=6...... any help appreciated!

#2 Ally

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Posted 18 April 2005 - 10:05 AM

(x - 2) and (x + 3) are factors of f(x) = 3x3 + 2x2 + cx + d

You can solve this question by using long division or by subbing in the values of x into f(x). The latter is the easiest way.

Since (x - 2) is a factor, (x - 2) = 0, so x = 2.
Since (x + 3) is a factor, (x + 3) = 0, so x = -3.

f(x) = 3x3 + 2x2 + cx + d
f(2) = 3(2)3 + 2(2)2 + c(2) + d
f(2) = 32 + 2c + d

f(-3) = 3(-3)3 + 2(-3)2 + c(-3) + d
f(-3) = -63 - 3c + d

32 + 2c + d = -63 - 3c + d
5c = -95
c = -19

Subbing in c= -19 into:

32 + 2(-19) + d = 0
- 6 + d= 0
d = 6

#3 kr88

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Posted 18 April 2005 - 10:55 AM

that was great i understand now.. thanks for taking the time 2 do those 2 Qs! ur immense!





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