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anonymous

  • 5 years ago

any help would be appreciated.. I have a test, and I am unsure of my probs below... ty.

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  1. anonymous
    • 5 years ago
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    where are the problems?

  2. anonymous
    • 5 years ago
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    I have one... Given that the polynomial f(x) has a zero at x=2 find the remaining zeros.. f(x)=x^3-10x^2+41x-50

  3. nowhereman
    • 5 years ago
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    You can factor (x-2) from f(x) if it has a zero at x=2.

  4. anonymous
    • 5 years ago
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    What does that mean? It says my answer should be in a+bi form?

  5. anonymous
    • 5 years ago
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    having a zero at x=2 means that (x-2) is one factor of f(x), you have to find the other factors.. using long division gives: \[f(x)=(x-2)(x^2-8x+25)=0\] you already have the solution x=2 to find the other solutions solve the quadratic equation: \[x^2-8x+25=0\]

  6. anonymous
    • 5 years ago
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    using the quadratic formula gives us the two remaining solutions: \[x=4-3i, x=4+3i\]

  7. anonymous
    • 5 years ago
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    do you know how to use the quadratic formula?

  8. anonymous
    • 5 years ago
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    Long division.. we are being taught syntheic.. is that ok?

  9. nowhereman
    • 5 years ago
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    there is only one solution to that division, and that step is really trivial it does not matter how you solve it.

  10. anonymous
    • 5 years ago
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    Can you show me the way the quadratic formula is worked out on here??

  11. nowhereman
    • 5 years ago
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    The solution formula for \[ x^2 + px + q = 0\] is \[x = -\frac{p}{2} \pm \sqrt{\frac{p^2}{4} - q}\]

  12. anonymous
    • 5 years ago
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    And thats the quadratic formula??

  13. anonymous
    • 5 years ago
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    becasue I have never seen that before??

  14. nowhereman
    • 5 years ago
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    Some schools teach the form \[ax^2 + bx +c = 0\] where you would of course get a formula which is a bit different. But the above is the exact solution formula for such quadratic problems.

  15. anonymous
    • 5 years ago
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    Thanks

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