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anonymous

  • 5 years ago

any idea how to get the zeros for x^3-4x^2+x+6 without using a graphing calculator? (can't figure out how to factor it out)

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  1. anonymous
    • 5 years ago
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    I should say algebraically, I know I can make a table and just get lucky

  2. dumbcow
    • 5 years ago
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    try to factor out (x+-1), (x+-2)...up to 6 if that works you'll be left with a quadratic

  3. dumbcow
    • 5 years ago
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    actually you can factor out (x+1) there are different ways of doing this (x+1)(x^2 +bx +6) = x^3-4x^2+x+6 we have to find b, if b is an integer it works x^2+bx^2 = -4x^2 (b+1)x^2 = -4x^2 b+1 = -4 b = -5

  4. anonymous
    • 5 years ago
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    ah long division, its been a while

  5. dumbcow
    • 5 years ago
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    long division always works too :)

  6. anonymous
    • 5 years ago
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    thanks

  7. anonymous
    • 5 years ago
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    You could also use the rational root theorem to discover the possible rational roots.

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