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| Question 57446This question is from textbook Applied College Algebra
 :  Find the real zeros of P(x) = x^4 - x^3 - 18x^2 + 52x - 40. If a zero is a multiple zero, state its multiplicity. 
This question is from textbook Applied College Algebra
 
 Answer by Nate(3500)
      (Show Source): 
You can put this solution on YOUR website! P(x) = x^4 - x^3 - 18x^2 + 52x - 40 Descartes Rule of Signs:
 f(x) ~> (+) ~> (-) ~> (-) ~> (+) ~> (-)
 3 changes ....
 f(-x) ~> (+) ~> (+) ~> (-) ~> (-) ~> (-)
 1 change ....
 Pos: 3 or 1
 Neg: 1 or 1
 Img: 0 or 2
 a = 1,2,4,5,8,10,20,40
 b = 1
 +- a/b = +-1,+-2,+-4,+-5,+-8,+-10,+-20,+-40
 We know for sure there is a negative zero:
 2|....1....-1....-18....52....-40
 ........1.....1....-16....20......0
 (x - 2)(x^3 + x^2 - 16x + 20)
 We know for sure there is a positive zero as well:
 -5|....1....1....-16....20
 .........1....-4.....4......0
 (x - 2)(x + 5)(x^2 - 4x + 4)
 (x - 2)(x + 5)(x - 2)(x - 2) = P(x)
 Zero Product Property:
 x = 2
 x = -5
 
  
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