C3 polynomials and the remainder theorem Watch

Core
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#1
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A questions asks me to devide  x^3 +x^2 -7 by  x-3
by using the remainder theorem, while reading the solution I am told, (this is after they find D, the remainder), to substitute  x=0 and D = 29
why did i have to make x = 0? The book dosnt explain this.

 x^3 + x^2 -7=(Ax^2 + Bx + C)(x -3) +D after solviing for D and substituting D value and x = 0  0+0-7= (A \cdot 0 + B \cdot 0 + C) \cdot (0 -3) +29
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Nicknak256
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You put x=0 to get rid of the x's so can work it out easier

i.e. -7 = -3C + 29
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Core
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(Original post by Nicknak256)
You put x=0 to get rid of the x's so can work it out easier

i.e. -7 = -3C + D
So i can just get rid of x? In which situations could i reuse this method?
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clad in armour
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how the hell is that C3?
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Nicknak256
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(Original post by Core)
So i can just get rid of x? In which situations could i reuse this method?
Yeah, since 0X is basically 0, since anything multiplied by 0 is 0. I can't really think now where you would reuse this method but if you are given two equations which equal each other, like here (and also chapter 1 of C4 if you're doing that too etc), and you chose which values of x to substitute into it to find the unknowns, most of the time it's easier to substitute x=0
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Mechie
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Could you not just use algebraic long division? http://en.wikipedia.org/wiki/Polynomial_long_division
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Core
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(Original post by davidmarsh01)
Could you not just use algebraic long division? http://en.wikipedia.org/wiki/Polynomial_long_division
Unfortunately it specificaly asked me to use the remainder theorem just as i have been asked to use the trapezium rule in a test paper when i could have easily difrentiated, i suppose hey want us to no alternative methods that can be used in different situations.
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Core
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(Original post by Nicknak256)
Yeah, since 0X is basically 0, since anything multiplied by 0 is 0. I can't really think now where you would reuse this method but if you are given two equations which equal each other, like here (and also chapter 1 of C4 if you're doing that too etc), and you chose which values of x to substitute into it to find the unknowns, most of the time it's easier to substitute x=0
I should have paid more attention to the question ty for your help.
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Nicknak256
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(Original post by Core)
I should have paid more attention to the question ty for your help.
No problem
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