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anonymous

  • 5 years ago

Factor x3 + 2x2 + 5x + 10

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  1. amistre64
    • 5 years ago
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    well this ones tricker, you have to do more of a trial and error to break apart the "^3" cubic part.

  2. amistre64
    • 5 years ago
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    we can increase our odds by taking the factors of 10 as a pool of numbers to test. 1,10; 2,5 I use synthetic division to test it becasue its just less clutter

  3. amistre64
    • 5 years ago
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    lets try x = -1: -1| 1 2 5 10 0 -1 -1 -4 ----------- 1 1 4 6 <- not a zero, not a factor

  4. amistre64
    • 5 years ago
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    x = -2 maybe? -2| 1 2 5 10 0 -2 0 -10 ----------- 1 0 5 0 <- zero!! we have a winner..

  5. amistre64
    • 5 years ago
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    (x+2)(x^2+5x)

  6. amistre64
    • 5 years ago
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    opps...silly superhero.... (x+2) (x^2+5)

  7. anonymous
    • 5 years ago
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    Try factoring by grouping. x^3 + 2x^2 + 5x + 10 x^2(x+2)+5(x+2) (x^2+5)(x+2)

  8. anonymous
    • 5 years ago
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    lol

  9. amistre64
    • 5 years ago
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    hey grouping does work with this one.... that was cool ;)

  10. amistre64
    • 5 years ago
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    of course its not as brainiacal as mine...but you know ;)

  11. anonymous
    • 5 years ago
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    yes lol

  12. anonymous
    • 5 years ago
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    9x2 + 48x + 64

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