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Let $f(x)=x^4-2x^3+3x^2-2x+1$. Prove that $f(x) \ge 0$ .

My thought is I need factor the function into sum or difference of perfect squares to show it's always non-negative. Any suggestions?

Drake
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    It is difficult to provide you with a good answer, as you have not provided us with much context. Why are you interested in this problem? Is it for a class? What class (high school algebra? college pre-calculus? calculus? analysis?)? What tools do you have available (memorized formulae for factoring polynomials? the derivative and Fermat's theorem? something else?)? Please edit your question to include some additional context. – Xander Henderson Jan 21 '19 at 21:39
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    I agree with @XanderHenderson. To test your intuition, can you see that it is positive if $x$ is negative? – Mark Bennet Jan 21 '19 at 21:50

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It's just $$(x^2-x+1)^2>0$$ Also, we can use your idea: $$f(x)=x^4-2x^3+x^2+x^2+x^2-2x+1=x^2(x-1)^2+x^2+(x-1)^2>0.$$

Michael Rozenberg
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  • @jordan_glen I added something for you. Also, for you: I just solved during my life one problem or maybe two. There for, it seems to you like magic. – Michael Rozenberg Jan 21 '19 at 21:46
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    @jordan_glen Your message is clear and I agree with it, however your delivery, particularly the sardonicism attached to it, is extremely unwarranted and unnecessary. – Rhys Hughes Jan 21 '19 at 21:48
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    Strictly the question is $\ge 0$, so $\gt 0$ is a bonus. The OP seems to be looking for help to see how to solve the question when it doesn't look obvious. My focus was on the $-2$ coefficients which I know belong to simple squares. There is a difference between solving a problem, and coaching someone else to solve it for themselves. I enjoy solving the problems on the site a lot, but I enjoy most the answers which help others to solve the problems too. – Mark Bennet Jan 21 '19 at 21:54