Step 1: Factor \(2 x^{2} - 3x -2\)
1. Multiply 2 by -2, which is -4.
2. Ask: Which two numbers add up to -3 and multiply to -4?
3. Answer: 1 and -4
4. Rewrite \(-3x\) as the sum of \(x\) and \(-4x\)
\(2 x^{2} +x-4x-2\ \textless \ 0\)
Step 2: Factor out common terms in the first two terms, then in the last two terms.
\(x(2x+1)-2(2x+1)\ \textless \ 0\)
Step 3: Factor out the common term \(2x+1\)
\((2x+1)(x-2)\ \textless \ 0\)
Step 4: Solve for \(x\)
1. Ask: When will \((2x+1)(x-2)\) equal zero?
2. Answer: When \(2x + 1 = 0\) or \(x-2=0\)
3. Solve each of the 2 equations above:
\(x=- \frac{1}{2} ,2\)
Step 5: From the values of \(x\) above, we have these 3 intervals to test.
x = < -1/2
-1/2 < x < 2
x > 2
Step 6: Pick a test point for each interval
For the interval \(x\ \textless \ - \frac{1}{2}\)
Lets pick \(x=-1.\) Then, \(2(-1) ^{2} -3 * -1 -2 \ \textless \ 0\)
After simplifying, we get \(3\ \textless \ 0\), Which is false.
Drop this interval.
For this interval \(- \frac{1}{2} \ \textless \ x\ \textless \ 2\)
Lets pick \(x=0\). Then, \(2* 0^{2} - 3 * 0-2\ \textless \ 0\). After simplifying, we get \(-2\ \textless \ 0,\) which is true. Keep this interval.
For the interval \(x\ \textgreater \ 2\)
Lets pick \(x = 3.\) Then, \(2 * 3 ^{2} -3*3-2\ \textless \ 0.\) After simplifying, we get \(7\ \textless \ 0\), Which is false. Drop this interval.
.Step 7: Therefore,
\(- \frac{1}{2} \ \textless \ x\ \textless \ 2\)
Done! :)
