anonymous
  • anonymous
how can you find the roots of this polynomial equation:2x^4-7x^#+21x^2+17x-13?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
That's a monster!
anonymous
  • anonymous
First check if \(2\) is a root.
anonymous
  • anonymous
Check \(1, -1, 2, -2\) and pray one of them is a root.

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More answers

anonymous
  • anonymous
If one of them is a root then divide out \(x-\text{root}\)
wolfe8
  • wolfe8
I think you're supposed to use factorization.
anonymous
  • anonymous
If you see how to do that @wolfgirl then show how.
anonymous
  • anonymous
i did use syntethetic division and use -1 and 1/2 as the divisor then after that i get the answer 2x^2-8x+26 then since i couldny use factorixation then i use quadratic formula and my answers turn out to be 10sqrtof11all over 4 and 6sqrt of 11 all over 4
anonymous
  • anonymous
and the problem is idki if thats was correct ill appreciate if youll tell me my errors
anonymous
  • anonymous
Show your work
wolfe8
  • wolfe8
Square both sides: 2 x^4-7 x^3+21 x^2+17 x-13 = 0 The left hand side factors into a product with three terms: (x+1) (2 x-1) (x^2-4 x+13) = 0 I think you can do the rest. Btw, I'm not Wolfgirl :)
wolfe8
  • wolfe8
Also, you made a mistake in the question with that #. I know from chat though you meant 3
tkhunny
  • tkhunny
Don't check 2 or -2. The Rational Root Theorem suggests +/- 1, +/- 1/2, +/- 13, and +/- 13/2 By your result, you had hits on x = -1 and x = 1/2.
anonymous
  • anonymous
@tkhunny so is my results correct?
tkhunny
  • tkhunny
Typically, I would prefer (2x-1) in the place of (x-1/2), but other than that it seems fine.

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