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# f(x) = x^5 - 2x^4 - 2x^3 + 4x^2 - 3x + 3 has no rational roots.  Use the graphing...

kristenmarieb... | Student, Grade 10 | Valedictorian

Posted July 30, 2013 at 5:35 AM via web

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f(x) = x^5 - 2x^4 - 2x^3 + 4x^2 - 3x + 3 has no rational roots.  Use the graphing calculator to approximate the irrational solutions correct to the 3 decimals.  If there is more than 1 real solution, enter them from smallest to largest.

Tagged with algebra 2, math

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted July 30, 2013 at 5:42 AM (Answer #1)

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Find the real zeros for `f(x)=x^5-2x^4-2x^3+4x^2-3x+3` :

Graph the function:

Since the function is a quintic, it can have at most 4 turning points. From the graph there are 3 real zeros -- between -2 and -1,between 1 and 2, and between 2 and 3.

Using the calculate zero function we get:

`x~~-1.668496`

`x~~1.208171`

`x~~2.2441101`

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The real zeros to 3 decimal places are x=-1.668,x=1.208, x=2.244

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