How many real roots does $x^7+14x^5+16x^3+30x-560=0$ have?

Question

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).

Answer 1

Hint: $(x^7+14x^5+16x^3+30x-560)'=7x^6+70x^4+48x^2+30>0$ for any $x\in \mathbb R$.

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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).

Answer 2

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: $$\frac{\mbox{d}}{\mbox{d}x}\left( x^7+14x^5+16x^3+30x-560 \right)=7x^6+70x^4+48x^2+30>0$$

Answer 3

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

Answer 4

Hint: Number of real roots:

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