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How many real roots does $x^7+14x^5+16x^3+30x-560=0$ have?
(a) 1
(b) 3
(c) 5
(d) 7

I don't know what approach to use in order to solve this equation. All I know is that the highest degree of this equation is odd hence it is obvious that it has at least 1 real root. But anyways all options are $\geq 1$. It would be great if I could get a hint as to how I can approach this question.

Edit: I appreciate the answers I have received, but answers not including calculus would also be encouraged as the source for this question is the chapter "quadratic equations"(even though this equation is not really quadratic).

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    How about a sign analysis? – 2017-01-05
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    @Vincent for that I would have to factor this equation and I don't think that's an easy task – 2017-01-05
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    [Interesting read](http://www.mathsisfun.com/algebra/polynomials-rule-signs.html) – 2017-01-05

4 Answers 4

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Hint: Number of real roots:

Number of sign changes in $f(x)$ + Number of sign changes in $f(-x)$

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    $f(x)=x^7+14x^5+16x^3+30x-560$ has one change of sign, $f(-x)=-x^7+0x^6-14x^5+0x^5-16x^3+0x^2-30x-560$ has 6 change of sign. right? – 2017-11-03
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    No, your counter example is not correct. You have put $+$, and $-$ when you want. But, these are not fixed, because you can also put $-$, and $+$, if you don't want. – 2017-11-03
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    I got it! Coefficients=zero doesn't count. Am I right? – 2017-11-03
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    It has one positive real and 6 complex roots. Right? – 2017-11-03
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    @ManeeshNarayanan , yes. – 2017-11-06
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Have you worked with derivatives?

Notice that for $x=0$, the polynomial on the left-hand side is clearly negative. On the other hand, for $x$ sufficiently large, it is clearly positive. It has at least one real root, but that was suggested by the possible answers as well.

To see it cannot have more roots, notice that the derivative of that polynomial, a sum of positive terms, is positive, so it is (monotonically) increasing: $$\frac{\mbox{d}}{\mbox{d}x}\left( x^7+14x^5+16x^3+30x-560 \right)=7x^6+70x^4+48x^2+30>0$$

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    Thank you so much :) By any chance do you know any other way of solving this question which does not include calculus? – 2017-01-05
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    The other answers also refer to rules concerning the sign changes of the polynomial coefficients. You could use that, if you are allowed to use those rules. – 2017-01-05
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Hint: $(x^7+14x^5+16x^3+30x-560)'=7x^6+70x^4+48x^2+30>0$ for any $x\in \mathbb R$.

Another way, without derivative is to use Descartes' rule of signs. It says that the number of positive roots of the polynomial is at most equal to the number of sign differences between consecutive nonzero coefficients (in descending variable exponent). Here $$x^7+14x^5+16x^3\underbrace{+30x-560}_{\text{change from $+$ to $-$}}$$ hence you have at most one positive real root. Similarly, the rule applies again to give that you have at most one negative real root. Hence, you have at most two real roots which allows only one possible answer. The existence of one real root is immediate (you already have it).

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    hint ( hint ) : strictly increasing function – 2017-01-05
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    @Jimmy, you should not edit/add other's answer after an answer. This is bad habit on SE. – 2017-01-05
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    @MithleshUpadhyay Yes, I know, but would it be better to add a second answer in this case, or just leave it? What do you think? Your answer, does not mention that this is Descartes' rule of signs, so it should still be somewhere IMHO (by the way you got +1 from me). – 2017-01-05
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    @MithleshUpadhyay Not to be confused: I did not just rewrite your answer, I was working independently to found a way without derivative and it just took me some more time than the derivative. – 2017-01-05
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let $f(x) = x^7+14x^5+16x^3+30x-560=0$

then $f'(x) = 7x^6+70x^{4}+48x^2+30 >0$ for all real $x$

means function $f(x)$ is strictly increasing function and

when $x\rightarrow -\infty\;,$ then $f(x)\rightarrow -\infty$

when $x\rightarrow \infty\;,$ then $f(x)\rightarrow \infty$

so it will cut $x$ axis exactly at one point

So $f(x) = 0$ has exactly one real root