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!

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# 2002 Winter Diet, Paper 1, Q5

Started by kr88, Apr 18 2005 08:44 AM

2 replies to this topic

### #1

Posted 18 April 2005 - 08:44 AM

### #2

Posted 18 April 2005 - 10:05 AM

(x - 2) and (x + 3) are factors of f(x) = 3x

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) = 3x

f(2) = 3(2)

f(2) = 32 + 2c + d

f(-3) = 3(-3)

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

32 + 2c + d = -63 - 3c + d

5c = -95

Subbing in c= -19 into:

32 + 2(-19) + d = 0

- 6 + d= 0

^{3}+ 2x^{2}+ cx + dYou 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) = 3x

^{3}+ 2x^{2}+ cx + df(2) = 3(2)

^{3}+ 2(2)^{2}+ c(2) + df(2) = 32 + 2c + d

f(-3) = 3(-3)

^{3}+ 2(-3)^{2}+ c(-3) + df(-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**
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### #3

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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