TheKevinFang
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#1
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#1
Not why with question 14 (d), why the answer is square root of 5 (negative/positive), as well as -1.

Quick help is appreciated.
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poppy77
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If f(x)= (x+1), x^3 +x^2 - 4x - 4 = x+1;
Then x^3 + x^2 - 5x- 5 = 0
Factorising to (x^2-5)(x+1)
Hope this helps.
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TheKevinFang
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(Original post by poppy77)
If f(x)= (x+1), x^3 +x^2 - 4x - 4 = x+1;
Then x^3 + x^2 - 5x- 5 = 0
Factorising to (x^2-5)(x+1)
Hope this helps.
Thanks. Could you also help me with my part c?
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Dolphins
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Never mind you got it D:
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poppy77
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(Original post by Dolphins)
Never mind you got it D:
Haha I was doubting my subtraction skills there! No worries
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poppy77
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(Original post by TheKevinFang)
Thanks. Could you also help me with my part c?
Sure. (x-2)(x+1) is x^2 -x -2
x^3 + x^2 - 4x - 4 = x^2 -x - 2
So you just factorise x^3 - 3x - 2 into (x+1)^2(x-2)
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TheKevinFang
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(Original post by poppy77)
Sure. (x-2)(x+1) is x^2 -x -2
x^3 + x^2 - 4x - 4 = x^2 -x - 2
So you just factorise x^3 - 3x - 2 into (x+1)^2(x-2)
For part (d) why can't I do it this way? (attached)

I get the wrong answer, can I not factorise the original f(x) first?

Need a quick reply, kind regards.
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poppy77
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No, that doesn't work. If you wanted to divide through by (x+1), 1 would be left on the right hand side so the left would simplify to x^2 -5. It's much easier just to move all the terms to the left hand and set it equal to zero.
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