Review question

# When is $(1-x)^n (2-x)^{2n} (3-x)^{3n} (4-x)^{4n} (5-x)^{5n}$ negative? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6674

## Solution

If $x$ and $n$ are integers then $(1-x)^n (2-x)^{2n} (3-x)^{3n} (4-x)^{4n} (5-x)^{5n}$ is

1. negative when $n>5$ and $x<5$,

2. negative when $n$ is odd and $x>5$,

3. negative when $n$ is a multiple of $3$ and $x>5$,

4. negative when $n$ is even and $x<5$.

For convenience, let $f_n(x) = (1-x)^n (2-x)^{2n} (3-x)^{3n} (4-x)^{4n} (5-x)^{5n}.$

First consider the case $x=4, n = 6$. We have $f_6(4)=0$, which means that (a) and (d) are false.

Now consider the case $x=6, n=6$. Then each of $(1-x)^n$, $(2-x)^{2n}$, $(3-x)^{3n}$, $(4-x)^{4n}$ and $(5-x)^{5n}$ is greater than or equal to zero, which means that (c) is false.

It only remains to show that (b) is true. If $x>5$, then each of $1-x$, $2-x$, $3-x$, $4-x$ and $5-x$ is negative.

If $n$ is odd here then \begin{align*} (1-x)^n &<0\\ (2-x)^{2n}&>0\\ (3-x)^{3n}&<0\\ (4-x)^{4n}&>0\\ (5-x)^{5n}&<0 \end{align*}

so their product is negative, as required. The answer is (b).