Question

Factorise 2x^2 - 3x -2 < 0

Answer 1

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! :)