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| Question 656264:  ``the function f(x)=x(cubed)-2x(squared)-5x+6 has three integer roots
 
 1) Find the tree roots?
 2)Find a cubic euation wose roots are 1 less than the roots of f.
 
 PLEASE HELP ME''.
 
 Answer by jim_thompson5910(35256)
      (Show Source): 
You can put this solution on YOUR website! Factors of 6: 1, 2, 3, 6, -1, -2, -3, -6 
 Factors of 1: 1, -1
 
 *** Note: Include negative factors as well ***
 
 Divide all the factors of 6 by the factors of 1 to get the potential rational roots of x^3-2x^2-5x+6:
 
 1, -1, 2, -2, 3, -3, 6, -6
 
 Note: there are 8 possible rational roots (listed above) for x^3-2x^2-5x+6.
 
 
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 Let's check all the possible roots to see if there are any actual rational roots.
 
 
 Check to see if x = 1 is a root for x^3-2x^2-5x+6:
 
 (1)^3-2(1)^2-5(1)+6 = 0
 
 Since the result is 0, x = 1 is a root for x^3-2x^2-5x+6.
 
 
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 Check to see if x = -1 is a root for x^3-2x^2-5x+6:
 
 (-1)^3-2(-1)^2-5(-1)+6 = 8
 
 Since the result is NOT 0, x = -1 is NOT a root for x^3-2x^2-5x+6.
 
 
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 Check to see if x = 2 is a root for x^3-2x^2-5x+6:
 
 (2)^3-2(2)^2-5(2)+6 = -4
 
 Since the result is NOT 0, x = 2 is NOT a root for x^3-2x^2-5x+6.
 
 
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 Check to see if x = -2 is a root for x^3-2x^2-5x+6:
 
 (-2)^3-2(-2)^2-5(-2)+6 = 0
 
 Since the result is 0, x = -2 is a root for x^3-2x^2-5x+6.
 
 
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 Check to see if x = 3 is a root for x^3-2x^2-5x+6:
 
 (3)^3-2(3)^2-5(3)+6 = 0
 
 Since the result is 0, x = 3 is a root for x^3-2x^2-5x+6.
 
 
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 Check to see if x = -3 is a root for x^3-2x^2-5x+6:
 
 (-3)^3-2(-3)^2-5(-3)+6 = -24
 
 Since the result is NOT 0, x = -3 is NOT a root for x^3-2x^2-5x+6.
 
 
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 Check to see if x = 6 is a root for x^3-2x^2-5x+6:
 
 (6)^3-2(6)^2-5(6)+6 = 120
 
 Since the result is NOT 0, x = 6 is NOT a root for x^3-2x^2-5x+6.
 
 
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 Check to see if x = -6 is a root for x^3-2x^2-5x+6:
 
 (-6)^3-2(-6)^2-5(-6)+6 = -252
 
 Since the result is NOT 0, x = -6 is NOT a root for x^3-2x^2-5x+6.
 
 
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 Answer:
 
 The polynomial x^3-2x^2-5x+6 has 3 rational roots and they are: x = 1, x = -2, x = 3,
 
 So the rational factors are: (x - 1), (x + 2) and (x - 3)
 
 This means that x^3-2x^2-5x+6 factors to (x - 1)(x + 2)(x - 3)
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