MHB Find x- and y- Intercepts....4

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The discussion focuses on finding the x- and y-intercepts of the polynomial y = 2x^5 - 5x^4 + 5. The y-intercept is determined to be at the point (0,5) when x is set to 0. To find the x-intercept, the equation 0 = 2x^5 - 5x^4 + 5 is analyzed, but it is noted that this polynomial likely does not have roots expressible in elementary functions. The complexity of solving quintic equations is acknowledged, and the use of numerical root-finding techniques or tools like Wolfram Alpha is recommended for approximating real roots. The discussion concludes with an understanding that advanced techniques may be necessary for finding "nice" roots in this case.
mathdad
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Find the x- and y-intercepts.

y = 2x^5 - 5x^4 + 5

Let x = 0

y = 2(0)^5 - 5(0)^4 + 5

y = 5

The y-intercept is y = 5 and it takes place at the point (0,5).

Let y = 0 to find the x-intercept.

0 = 2x^5 - 5x^4 + 5

I am stuck here.
 
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I don't think $0 = 2x^5 - 5x^4 + 5$ has roots that can be expressed in terms of elementary functions.
 
More "nice" roots, right?
 
There's always a solution(s) for polynomials of degree 4 and lower (if one includes complex-valued results). Once we reach a degree of 5 we are, in general, out of luck (so to speak) but there are some quintics (polynomials of degree 5) that are solvable. I don't think the one we have here is, though.

A useful online tool is W|A (Wolfram Alpha). I will suggest that you familiarize yourself with this tool as it can aid insight into problems that may be difficult. :D
 
For any polynomial of degree 3 or more that doesn't succumb to the rational roots theorem to get down to a combination of linear/quadratic factors, I rely on numeric root finding techniques to approximate the real roots (and those are the only roots we need for finding intercepts). :D
 
You are saying that "nice" roots in this case require advanced techniques. I am not there yet. I get it now.
 
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