# Thread: More polynomial inequalities question

1. ## More polynomial inequalities question

I had another question for the problem (x(x-8))/(x^2-9x-52)<=0

my work is:
x(x-8)/(x+4)(x-13)<=0

i then did a number line with the numbers -4,0,8,13 and did the sign chart underneath. The answer I ended getting is (-4,0)U(8,13) I retried the problem several times and still got the same answer.

I also did the problem (x-4)/(x+13)=>0 with that i got the answer (-infinity,-13]U[4,infinity)

2. The answer for the first one you have is almost correct. It is $\left( { - 4,0} \right]\bigcup {\left[ {8,13} \right)}$ rather than $\left( { - 4,0} \right)\bigcup {\left( {8,13} \right)}$.

Same deal with the second. It is $\left( { - \infty ,13} \right)\bigcup {\left[ {4,\infty } \right)}$ not $\left( { - \infty ,13} \right]\bigcup {\left[ {4,\infty } \right)}$.

Those intervals can be tricky at times. If it is undefined at the endpoint, you should use the open interval. Also, always use an open interval for $\infty$, as you seem to know.

Edit: forgot to add, the interval is closed in the first problem at 0 and 8 because those values make the function 0, not undefined. Since the inequality wants 0 included, the interval is closed. Hopefully that clarifies it a bit.

3. 2. (-infinity,-13)U[4,infinity)

4. The key idea to solving a problem like $\frac{x(x-8)}{(x+4)(x-13)} < 0$ is to use the fact that the numerator and denominator must have different signs. So either

$x(x-8) > 0$ AND $(x+4)(x-13) < 0$

OR

$x(x-8) < 0$ AND $(x+4)(x-13) > 0$.

Let's look at the first case:

Question: When is $x(x+8) > 0$?
Answer: When $x$ and $x+8$ are both positive OR both negative.

$x > 0$ AND $x-8 > 0 \Rightarrow x > 8. \text{ } (8,\infty)$
OR
$x < 0$ and $x - 8 < 0 \Rightarrow x < 0. \text{ }(-\infty,0)$

Conclusion 1:
Putting this together we say the numerator is positive when $x$ is in $(-\infty,0) \cup (8,\infty)$

Question: When is $(x+4)(x-13) < 0$?
Answer: When $x+4$ and $x-13$ have opposite signs.

This happens when:
$x < -4$ AND $x > 13$(this is not possible)
OR
$x > -4$ AND $x < 13$. $\text{ } (-4,13)$

Conclusion 2:
So we conclude that the denominator is negative when $x$ is in $(-4,13)$.

Finally, we require that the numerator is positive AND the denominator is negative meaning both the first and second conclusions are met. This happens only when $x$ is in both intervals (their intersection) yielding as you pointed out $(-4,0) \cup (8,13)$. Now check when $\frac{x(x-8)}{(x+4)(x-13)} = 0$ and also check the other case: $x(x-8) < 0$ AND $(x+4)(x-13) > 0$.

Whenever you have an OR you are talking union:

$x$ is in $A$ OR $B$ $\iff x$ is in $A \cup B.$

Whenever you have an AND you are talking intersection:

$x$ is in $A$ AND $B$ $\iff x$ is in $A \cap B.$