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Qwerty90

  • one year ago

Question about how to solve: Determine the zeros of f(x) = x4 – x3 + 7x2 – 9x – 18. I am solving this problem using all of the polynomial ways such as decartes rule, fundamental therom and rational root. But it seems when I solve it that way, I either come up one answer short or a completely different answer from what I see everyone else get. How would you solve this and what is the answer?

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  1. ZeHanz
    • one year ago
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    RRT says: Rational solutions can only be factors of 18 divided by factors of the coefficient of x^4. Rational roots therefore have the form:\[\pm 18, \pm9, \pm6, \pm3, \pm2, \pm1\]If you try (begin with the simplest, of course, so 1 or -1), you get -1 as a root. You can now write f(x) as (x+1)(x³ ..........). To get the 3rd degree part, you can do a long division or a synthetic division. As soon as you have found it, you can do this trick again: I think 2 is a root as well (just try it), so you can factor out x-2 to get (x+1)(x-2)(x²........). Now you can esily check if there are more rational or even real roots.

  2. ZeHanz
    • one year ago
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    Could it be that you are not familiar with synthetic division? It's a real handy trick!

  3. Qwerty90
    • one year ago
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    No I know how to do synthetic division, its just after that where everything get messy.

  4. ZeHanz
    • one year ago
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    OK, so you've found: f(x)=(x+1)(x-2)(x²+9). As you can see, the last factor x²+9 has no real roots, only complex ones: 3i and -3i.

  5. Qwerty90
    • one year ago
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    So, basically after you found which zeros is part of the function you list it in parentheses as the conjugate. Now my last question is where did you get the x+9 from? i know that 9 is the constant but what about the x^2?

  6. ZeHanz
    • one year ago
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    After the first synthetic division, I got (x+1)(x³-2x²+9x-18). Then the number 2 is also a zero, so I divided x³-2x²+9x-18 by x-2. That gave me x²+9

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