SOLUTION: 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.

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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