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anonymous

  • one year ago

Find a cubic function with the given zeros.

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  1. anonymous
    • one year ago
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  2. mathstudent55
    • one year ago
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    A polynomial equation with roots a, b, c is f(x) = (x - a)(x - b)(x - c)

  3. anonymous
    • one year ago
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    f(x) = (x - a)(x - b)(x - c)\[f(x) = (x - \sqrt{2})(x - \sqrt{-2})(x - (-2))\]

  4. anonymous
    • one year ago
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    ?? @mathstudent55

  5. anonymous
    • one year ago
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    ugh im so confused

  6. mathstudent55
    • one year ago
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    You are close, but a little off. This is what you should have: f(x) = (x - a)(x - b)(x - c) \(f(x) = (x - \sqrt{2})(x - (-\sqrt{2}))(x - (-2))\) \(f(x) = (x - \sqrt{2})(x +\sqrt{2})(x + 2)\) Now multiply out the three binomials. Hint: Make sure to multiply the first two binomials first because since they are the product of a sum and a difference, you can use the short cut: \((a + b)(a - b) = a^2 - b^2\) instead of having to use FOIL.

  7. anonymous
    • one year ago
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    f(x)=(x−√2)(x+√2)(x+2) f(x)=2-2+2?

  8. anonymous
    • one year ago
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    its either A f(x) = x^3 + 2x^2 - 2x + 4 B f(x) = x^3 + 2x^2 + 2x - 4 C f(x) = x^3 - 2x^2 - 2x - 4 D f(x) = x^3 + 2x^2 - 2x - 4

  9. anonymous
    • one year ago
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    @Hero i need help too please

  10. anonymous
    • one year ago
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    @mathstudent55

  11. anonymous
    • one year ago
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    (a + b^)3 = (a + b)(a^2 + 2ab + b^2) = a^3 + 3a^2b + 3ab^2 + b^3??

  12. mathstudent55
    • one year ago
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    \(f(x) = \color{red}{(x - \sqrt{2})(x +\sqrt{2})}(x + 2)\) becomes \(f(x) = \color{red}{(x^2 - 2)}(x + 2)\) Ok so far?

  13. anonymous
    • one year ago
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    ok

  14. mathstudent55
    • one year ago
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    Now we use FOIL for the two binomials above: \(f(x) = (x^2 - 2)(x + 2)\) \(f(x) = x^3 + 2x^2 - 2x - 4\)

  15. anonymous
    • one year ago
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    ohhh ok

  16. anonymous
    • one year ago
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    so i do like

  17. anonymous
    • one year ago
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    x^2*x and so on

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