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How can I modify 0=x^4(x-43)+40x^2(18x-145)+16(1375x-1884) further?  I do not want to expand it, instead I want to shorten it, if you know what I mean so I can find the zeros. Please give step by step instructions.

 May 16, 2015

Best Answer 

 #1
avatar+118587 
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hi Gibsonj338,

 

$$0=x^4(x-43)+40x^2(18x-145)+16(1375x-1884)$$

I do not know any way to do this without expanding first and then looking for factors

in this case (x-12) is a factor.

12 is the only real solution.  There are 2 irrational solutions and 2 complex solutions

 

 

http://www.wolframalpha.com/input/?i=0%3Dx%5E4%28x-43%29%2B40x%5E2%2818x-145%29%2B16%281375x-1884%29+#

 May 17, 2015
 #1
avatar+118587 
+5
Best Answer

hi Gibsonj338,

 

$$0=x^4(x-43)+40x^2(18x-145)+16(1375x-1884)$$

I do not know any way to do this without expanding first and then looking for factors

in this case (x-12) is a factor.

12 is the only real solution.  There are 2 irrational solutions and 2 complex solutions

 

 

http://www.wolframalpha.com/input/?i=0%3Dx%5E4%28x-43%29%2B40x%5E2%2818x-145%29%2B16%281375x-1884%29+#

Melody May 17, 2015

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