Could you explain it to me thoroughly please
So far, I have:
-2/7<x<1
or
-2.5<x<1
or
x>1
Am I wrong?
set each factor equal to 0 and solve for x ...Solve this polynomial inequality algebraically (number line)? (7x+2) (1-x)(2x-5)>0
you should get $\displaystyle x = -\frac{2}{7}$ , $\displaystyle x = 1$ , and $\displaystyle x = \frac{5}{2}$
plot these three x-values on a number line in correct order.
the three plotted values of x break up the number line into four intervals ...
$\displaystyle x < -\frac{2}{7}$
$\displaystyle -\frac{2}{7} < x < 1$
$\displaystyle 1 < x < \frac{5}{2}$
$\displaystyle x > \frac{5}{2}$
pick any single value in each interval, substitute it into the original inequality, and see if the result makes the original inequality true or false.
if true, then all values in that interval will make the inequality true ... the interval is part of the solution set.
if false, reject that interval as part of the solution set.
A quick way to do that is to recognize that x- a is negative if x< a, positive if x> a. In (7x+2) (1-x)(2x-5)= (7x+2)(-1)(x-1)(2x-5)>0 there are three factors which are 0 at x= -2/7, x= 1, and x= 5/2 and a single (-1) factor. Notice that I changed 1-x to (1)(x- 1) to get the form "x-a".
Start with x< -2/7. Then x is less than all three of those numbers so all 3 factors are negative and we have (-1) which is always negative for 4 negative factors. Since there are an even number of negatives, the product is positive.
Now jump past -2/7: -2/7< x< 1. The factor 7x+2 changes to positive while the other three factors remain negative. Since there are an odd number of negatives, the product is negative.
Jump past 1 to 1< x< 5/2. The factor x-1 changes to positive so now we have two negative factors. Since there are an even number of negatives, the product is positive.
Finally jump past 5/2 to 5/2< x. Now all the factors except (-1) are positive so we have only one negative factor. Since there are an odd number of negatives, the product is negative.
Summary: the product is positive for x< -2/7 and for 1< x< 5/2.
This works nicely when the function is a polynomial that can be factored or a rational function in which the numerator and denominator can both be factored. Skeeter's method, choose one point in each interval, will work for all kinds of functions.
one more time ...
$\displaystyle (7x+2) (1-x) (2x+5) > 0$
1. set each factor equal to 0 and solve for x.
this time, you should get $\displaystyle x = -\frac{2}{7}$ , $\displaystyle x = 1$ , and $\displaystyle x = -\frac{5}{2}$
2. plot these three x-values on a number line in correct order.
the three plotted values of x break up the number line into four intervals ...
$\displaystyle x < -\frac{5}{2}$
$\displaystyle -\frac{5}{2} < x < -\frac{2}{7}$
$\displaystyle -\frac{2}{7} < x < 1$
$\displaystyle x > 1$
3. pick any single value in each interval, substitute it into the original inequality, and see if the result makes the original inequality true or false.
if true, then all values in that interval will make the inequality true ... the interval is part of the solution set.
if false, reject that interval as part of the solution set.
follow the above directions carefully, and you will find the solution set.
A quick way on solving inequalities like this one is the following: locate critical points then pick every factor and construct the following table,
$\displaystyle \begin{array}{*{20}c}
{}&\vline &{\left( {-\infty ,-\dfrac{5}
{2}} \right)} &\vline & {\left( {-\dfrac{5}
{2},-\dfrac{2}
{7}} \right)} &\vline & {\left( {-\dfrac{2}
{7},1} \right)} &\vline & {(1,\infty )}\\
\hline
{7x + 2}&\vline&-&\vline &-&\vline &+&\vline&+\\
\hline
{1 - x}&\vline&+&\vline &+&\vline &+&\vline&-\\
\hline
{2x + 5}&\vline&-&\vline&+&\vline&+&\vline&+\\
\hline{}&\vline&+&\vline&-&\vline&+&\vline&-
\end{array}$
So, the solution is $\displaystyle S=\left( -\infty ,-\frac{5}{2} \right)\cup \left( -\frac{2}{7},1 \right).$