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=- (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 \ - (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
`- (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,

`- (1)/(2) \ \textless \ x\ \textless \ 2`

Done! :)