The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is
I don't know how to solve this.
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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!
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.
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").
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).