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# C1 question Watch

1. Hi there. How would I go about doing this question below...?

3. (Original post by Traceur)
Is the answer x^3 - 5x^2 + 3x + 9 ? If so i can help you.
4. This is pretty straightforward

Well you know that x =-1 and touches x at 3

Which means you can compile the cubic via the inverse.

(x+1) (x-3)(x-3)

Muliply those out and you'll get: x^3- 5x^2 +3x+9=0

As you can see clearly a=-5

got it?
5. (Original post by marek35)
This is pretty straightforward

Well you know that x =-1 and touches x at 3

Which means you can compile the cubic via the inverse.

(x+1) (x-3)(x-3)

Muliply those out and you'll get: x^3- 5x^2 +3x+9=0

As you can see clearly a=-5

got it?
Ah of course! Thank you!
6. (Original post by Traceur)
Ah of course! Thank you!
no problem

7. I'm pretty sure you could also have done it with simultaneous equations as well, although it's slightly more strenuous and long winded! :P
8. Thanks for your answer but it surely seems to be much easier to use mareks solution...
I did think to use simultaneous equations but its unnecessarily complicated (for me anyway).
9. (Original post by Traceur)
Thanks for your answer but it surely seems to be much easier to use mareks solution...
I did think to use simultaneous equations but its unnecessarily complicated (for me anyway).
indeed solving it simultaneously would be pointless as you know where it crosses the x axis (or touches in this case too).

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