anonymous
  • anonymous
Find the value of a, of b and of c in the identity \[x^{3}+3x ^{2}-2x+16-c(x+2) = ax ^{2}(x-1)+b(x-2)^{2}(x-1)\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i don't see a snap way to do this sometimes it is easy by picking a nice value for \(x\) but i can't think of one here
anonymous
  • anonymous
you can either multiply all this junk out and equate like coefficients or you can replace \(x\) by say 0, 1, 2 and get three equations in \(a, b,c\)
anonymous
  • anonymous
I tried the method of comparing coefficients but I go an extremely weird answer and along the way had 6=0 :/

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anonymous
  • anonymous
put \(x=1\) first
anonymous
  • anonymous
you get \[1+3-2+16-3c=0\]
anonymous
  • anonymous
oh right, okay
anonymous
  • anonymous
i get \[c=6\] and then you can find exactly what the left side is

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