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This question already has an answer here:

Consider the polynomial $X^3+X^2+1 \in \mathbb R [X]$. Since it is of odd degree, it has at least one real root. How can I show that it's the only one?

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marked as duplicate by Dietrich Burde, user26857, Fly by Night, user228113, user147263 Apr 19 '16 at 23:07

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  • $\begingroup$ Root is when there is an equation . Here it would be zeroes as we assume the polynomial is equal to zero. Try to write the polynomial as a G.P $\endgroup$ – N.S.JOHN Apr 19 '16 at 15:06
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    $\begingroup$ I disagree. The statement of the fundamental theorem of algebra is that every nonconstant polynomial with complex coefficients has a complex root - there is no mention of an equation. $\endgroup$ – Showhat Apr 19 '16 at 15:08
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    $\begingroup$ Hint: study the derivative of $x\mapsto x^3+x^2+1$ and thus the behaviour of your function. $\endgroup$ – Raymond Manzoni Apr 19 '16 at 15:11
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    $\begingroup$ Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – Fly by Night Apr 19 '16 at 16:55
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    $\begingroup$ @TheGreatDuck I didn't say it wasn't. My comment makes my objection perfectly clear. $\endgroup$ – Fly by Night Apr 20 '16 at 14:47
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There are at least two possible approaches:

  1. Let $f(x)=x^3+x^2+1$. Differentiate $f$ to find the maxima and minima; $f$ will have three real roots iff the local maximum is above the $x$-axis and the local minimum is below the axis.

  2. If you want an algebraic proof, suppose for the sake of contradiction that the polynomial has real roots $a,b,c$. Then $abc=-1$ by Vieta's formulas, so at least one of $a,b,c$ has absolute value at least 1. But $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=1,$$ again by Vieta's, and this is a contradiction since none of $a,b,c$ is $0$.

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Derivative being equal to $0$ $$3x^2+2x=0$$ gives $x_1=0$ and $x_2=\frac{-2}{3}$ and in our initial polynomial we have $p(0)=1$ and $p(\frac{-2}{3})>0$ hence it never crosses the $0$ line again and cannot have any additional zeros amongst the real

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There is no positive root as all terms are positive. Zero is not a root. To check for negative roots, consider $-x^3+x^2+1$, which has only one sign change, so by Descartes rule of signs, has exactly one negative root.

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  • $\begingroup$ How can you say "zero is not a root". X = 0 could be a solution. $\endgroup$ – The Great Duck Apr 19 '16 at 17:56
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    $\begingroup$ @TheGreatDuck Not with a constant term $1$ added. $\endgroup$ – Macavity Apr 19 '16 at 17:57
  • $\begingroup$ You don't know what the expression equals. How can you say /anything/ about X itself? $\endgroup$ – The Great Duck Apr 19 '16 at 17:58
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    $\begingroup$ @TheGreatDuck When $x=0$ that expression is exactly $1$. That's all anyone needs to know to conclude it is not a root. Now unless you're going to make a valid point, bye. $\endgroup$ – Macavity Apr 19 '16 at 18:02
  • $\begingroup$ Sir, you are not answering the question at all. You are assuming the expression is equal to 0 which is complete BS. The person asked you to factor the expression, not solve for X. You cannot say that any number is a factor, and don't you tell me to go away. I have every right to ask why you are writing an answer that is completely inaccurate. $\endgroup$ – The Great Duck Apr 19 '16 at 18:15