anonymous
  • anonymous
1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
use rational root test which is a trial and error method. you use x's that are factors of the last term (-6) and the first term (4) so 1,-1,2,-2,4,-4 whichare the factors of 4 and 1,-1,2,-2,3,-3,6,-6 which are the factors of 6. So we try each one, see if it gives 0. Try 1, we see 4-5+9-6 is not 0 Try -1 we get 4(-1)^3 - 5(-1)^2 + 9(-1) + 6= -4 -5 -9 + 6 again not 0 Try 2 we get 4(2)^3 - 5(2)^2 + 9(2) - 6= 32-20+18-6 again not 0 Try -2 we get 4(-2)^3-5(-2)^2+9(-2)-6= -32-20 -18-6 which isnot 0 Try 3, we get 4(3)^3-5(3)^2+9(3)-6= 108-45+27-6 not 0 Try -3 again doesnt give 0 Try 4 again not 0 Try -4 again not 0 Try pluging all values till 6 and -6, it doenst look like this function have rational zeros.

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