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The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is

A. 0;
B. 3;
C. 5;
D. 1.

I don't know how to solve this.

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    $\begingroup$ consider factoring by grouping. $\endgroup$ – ECollins Mar 16 '16 at 18:57
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    $\begingroup$ Two things to consider when factoring polynomials over the reals: the rational root theorem and Descartes' rule of signs. Even if you cannot figure out how to factor it, it will give you an idea of how many roots there are and what they might be. $\endgroup$ – Cameron Williams Mar 16 '16 at 19:08
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As ECollins, Joanpemo, and peter.petrov, and GoodDeeds point out this equation is possible to factorize in more or less elementary terms which means it was possibly intended to be solved in that way.

However it is possible to use slightly stronger math (not necessarily available in the relevant course) to avoid part of that labour since the equation only has positive coefficients.

There is this delightful thing called Descartes' rule of signs which gives bounds on how many real roots an equation can have.It says that the number of real positive roots cannot exceed the number of sign changes in the coefficients.

Our equation has no sign changes at all so it cannot have any real positive roots!

Then you consider the polynomial $$ p(-x) = (-x)^5 + 2(-x)^3 + (-x)^2 + 2 = -x^5 - 2x^3 + x^2 + 2 $$

This polynomial only has one sign change in the coefficients so it can at most have one positive root meaning the original polynomial may at most have one real negative root.

Here we've already reduced ourselves to either A or D and if we're willing to chance it just random pick one of them.

However we don't need to do that because the polynomial in fact must have one real root. Look at it again.

$$p(x) = x^5 + 2 x^3 + x^2 + 2$$

It has odd degree! And polynomials with odd degrees always have at least one real root (which we in this case know to be negative) because it's limits at positive and negative infinity have different signs.

Therefore D (1) is the correct answer! and we didn't need to factor or solve for anything so know it!

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  • $\begingroup$ With all due respect: on one side you use a high school tool (Descartes' rule) to bound above the number of real roots, yet the job isn't complete until you use a tool from differential calculus: any real polynomial of odd degree has at least one real root, which is a straight application of the intermediate value theorem. Thus, Descartes isn't enough. Nice answer, though. $\endgroup$ – DonAntonio Mar 16 '16 at 19:30
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    $\begingroup$ I am not familiar the exact american curriculum as I reside in a nation that never on any level teaches the Descartes rule of signs while the odd-degree rule is taught early. But does the curriculum then usually contain the fact that if a polynomial has real coefficients then the (nonreal) complex roots come in conjugate pairs? In that case you could use the fact that the number of nonreal roots must be even, (0,2, or 4) which in either case demands at least one real root for them to total 5 in all (fundamental theorem of algebra). $\endgroup$ – Squid Mar 16 '16 at 19:38
  • $\begingroup$ Thank you. The fact that complex roots of real polynomials come in conjuugate pairs requires a basic knowledge of complex numbers. It is convered here in the highest level of mathematics in high school, yet the odd-degree rule cannot be properly taught without studying to some depth continuity, which doesn't happen (at least here) at HS level. Another thing is to state it an unproven rule, of course. $\endgroup$ – DonAntonio Mar 16 '16 at 19:43
  • $\begingroup$ How is the DRS properlytaught on a pre-calculus level? The proof I (barely) remember, while purely algebraic, is still pretty laborious to perform and the intuitive motivation that the roots (if any) follow from the different terms dominating at different scales causing the function to switch back and forth from increasing and decreasing depending on the signs is pretty much a calculus argument. $\endgroup$ – Squid Mar 16 '16 at 19:53
  • $\begingroup$ I don't really remember/know (I'm not sure if I ever knew), but the easiest proof I know uses some calculus. $\endgroup$ – DonAntonio Mar 16 '16 at 19:58
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This polynomial is equal to:

$(x^2+2)\cdot(x+1)\cdot(x^2-x+1)$

From here it's obvious that it has one real root only: $x=-1$.

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  • $\begingroup$ I accidentally downvoted this and only just noticed. If you make a minor typesetting edit I can change it to an upvote $\endgroup$ – Stella Biderman Mar 16 '16 at 20:04
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An idea following the comment to your question:

$$x^5+2x^3+x^2+2=x^3(x^2+2)+x^2+2=(x^3+1)(x^2+2)$$

Thus, the only real roots can come from $\;x^3+1\;$, and this has only one since

$$x^3+1=(x+1)(x^2-x+1)$$

and the above quadratic factor is irreducible.

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Observe by trial and error that $-1$ is a root. Then, $$x^5+2x^3+x^2+2=(x+1)(x^4-x^3+3x^2-2x+2)=(x+1)(x^4-x^3+x^2+2x^2-2x+2)=(x+1)(x^2(x^2-x+1)+2(x^2-x+1))=(x+1)(x^2+2)(x^2-x+1)$$

Thus, there is only one real root.

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    $\begingroup$ I would say trial and error is a poor technique in a more general situation. What if the polynomial had constant term 42 instead of 2? Edit: This is supposed to allude to the factor theorem. I forgot to mention this in the original comment! $\endgroup$ – Edward Evans Mar 16 '16 at 19:07
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If $x \ge 0$ the $x^5 + 2x^3 + x^2 + 2 \ge 2$ so there are no solutions for $x \ge 0$.

If $x = -humongouseffinglargenumber$ then $x^5 + 2x^3 + x^2+2 < 0$ so somewhere between $-humongouseffinglargenumber$ and $0$ there must be an $x$ where the term is $0$. Let's try to narrow that range down a little.

As we are only looking at negative values of $x$ we want to find values where $|x|^5 + 2|x|^3 = x^2 + 2$ and one such obvious place is when $|x| = 1$ i.e. when $x = -1$ thus $x^5 + 2x^3 + x^2 + 2 = -1 + -2 + 1 + 2 = 0$.

So that's one root. We can factor out $(x - (-1)) = x + 1$ to get $x^5 + 2x^3 + x^2 + 2 = (x + 1)(x^4 - x^3 + 3x^2 -2x + 2)$. So if there are any more real roots they are real roots to $x^4 - x^3 + 3x^2 - 2x + 2 = 0$.

As we are only considering negative $x$ these are the same as solving for $x^4 + |x|^3 + 3x^2 + 2|x| + 2 = 0$. As $|x| \ge 0$, $x^4 + |x|^3 + 3x^2 + 2|x| + 2 \ge 2$ so there are no solutions.

So $x = -1$ is the only solution.

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I highly recommend that you take advantage of computational knowledge engines when you're not sure how to proceed facing this type of problem. You can get a factorization from Symbolab ("Factor x^5 + 2x^3 + x^2 +2") and check your work by examining a plot on Wolfram Alpha ("factor x^5 + 2 x^3 + x^2 + 2").

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  • $\begingroup$ Using an engine to solve the problem is usually not an intention of the OP when OP asks a question. $\endgroup$ – Shailesh Mar 18 '16 at 7:35
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All the answers above are great and much more rigorous than what I am about to write; but it will be simpler to use the sign rule of Descartes- The max. no. of +ve real roots of a polynomial equation is the number of changes in sign of coefficients from +ve to -ve in f(x).

The max. no. of -ve real roots of a polynomial equation is the number of sign changes from -ve to +ve in f(-x).

Your equation f(x) has no change in the sign of coefficients( all are +ve). So no +ve real roots. If you write for f(-x), there is a change of sign in the $x^2$ coeffecient. Thus there is one -ve real root. SO the answer is (D); the equation has one real root (which is also negative).

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Consider the polynomial $\mod{x^3 +1} $

$x^5 + 2x^3 + x^2 + 2 \equiv -x^2 -2 + x^2 + 2 \equiv 0$

Thus, the polynomial has a factor of $x^3+1$

The remainder of the question can be done as above.

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