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Question 658208: x^3-x^2-7x+15 please state the total number of zeroes for the function both real and imaginary. using the rational root theorem.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Factors of 15: 1, 3, 5, 15, -1, -3, -5, -15
Factors of 1: 1, -1
*** Note: Include negative factors as well ***
Divide all the factors of 15 by the factors of 1
Potential rational roots of x^3-x^2-7x+15:
1, -1, 3, -3, 5, -5, 15, -15
Note: there are 8 possible rational roots for x^3-x^2-7x+15.
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Check to see if x = 1 is a root for x^3-x^2-7x+15:
(1)^3-(1)^2-7(1)+15 = 8
Since the result is NOT 0, x = 1 is NOT a root for x^3-x^2-7x+15.
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Check to see if x = -1 is a root for x^3-x^2-7x+15:
(-1)^3-(-1)^2-7(-1)+15 = 20
Since the result is NOT 0, x = -1 is NOT a root for x^3-x^2-7x+15.
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Check to see if x = 3 is a root for x^3-x^2-7x+15:
(3)^3-(3)^2-7(3)+15 = 12
Since the result is NOT 0, x = 3 is NOT a root for x^3-x^2-7x+15.
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Check to see if x = -3 is a root for x^3-x^2-7x+15:
(-3)^3-(-3)^2-7(-3)+15 = 0
Since the result is 0, x = -3 is a root for x^3-x^2-7x+15.
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Check to see if x = 5 is a root for x^3-x^2-7x+15:
(5)^3-(5)^2-7(5)+15 = 80
Since the result is NOT 0, x = 5 is NOT a root for x^3-x^2-7x+15.
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Check to see if x = -5 is a root for x^3-x^2-7x+15:
(-5)^3-(-5)^2-7(-5)+15 = -100
Since the result is NOT 0, x = -5 is NOT a root for x^3-x^2-7x+15.
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Check to see if x = 15 is a root for x^3-x^2-7x+15:
(15)^3-(15)^2-7(15)+15 = 3060
Since the result is NOT 0, x = 15 is NOT a root for x^3-x^2-7x+15.
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Check to see if x = -15 is a root for x^3-x^2-7x+15:
(-15)^3-(-15)^2-7(-15)+15 = -3480
Since the result is NOT 0, x = -15 is NOT a root for x^3-x^2-7x+15.
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So the polynomial x^3-x^2-7x+15 has exactly one rational root and it is x = -3.
So the only rational factor is (x + 3)
Since the degree is 3, there are 2 other roots. These two other roots are either irrational or complex roots.
Use polynomial long division, then use the quadratic formula to figure out what type of roots these other two roots are (I'll let you do this).
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