If $ N = \prod \limits_ {k=1}^{40} \bigl(x-(2k-1) \bigr)^{2k-1} $ then for how many integer values of $x$ is $N$ negative?
I am not getting any ideas for solving this one.
If $ N = \prod \limits_ {k=1}^{40} \bigl(x-(2k-1) \bigr)^{2k-1} $ then for how many integer values of $x$ is $N$ negative?
I am not getting any ideas for solving this one.
Hint: There is less to it than meets the eye. The powers are odd, so the sign of any term $(x-(2k-1))^{2k-1}$ is the same as the sign of $x-(2k-1)$. Our original polynomial, and the modified one where we replace each exponent by $1$, each have the roots $1$, $3$, $5$, and so on up to $79$. (You can keep the exponents if you like. It will make no real difference to the argument. But for me simplification is an automatic reflex.)
Look at a much smaller example, like $(x-1)(x-3)(x-5)(x-7)$. When $x<1$ or $x>7$, this is positive. The polynomial changes sign at each root. So we are negative at $2$, positive at $4$, negative at $6$, a total of $2$ integers at which we are negative. Now look at the same thing multiplied by $(x-9)(x-11)$. One more positive, one more negative, so there are $3$ integers for which our polynomial is negative. Now on to $40$ terms!