Question

How do you solve `(2x + 1) ( 4x ^ { 2} + 35x - 9) = 0`?

Answer 1

See the entire solution process below:

Explanation:

First, factor the binomial to:

`(2x + 1)(4x - 1)(x + 9) = 0`

Now, solve each term for `0`:

Solution 1)

`2x + 1 = 0`

`2x + 1 - color(red)(1) = 0 - color(red)(1)`

`2x + 0 = -1`

`2x = -1`

`(2x)/color(red)(2) = -1/color(red)(2)`

`(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -1/2`

`x = -1/2`

Solution 2)

`4x - 1 = 0`

`4x - 1 + color(red)(1) = 0 + color(red)(1)`

`4x + 0 = 1`

`4x = 1`

`(4x)/color(red)(4) = 1/color(red)(4)`

`(color(red)(cancel(color(black)(4)))x)/cancel(color(red)(4)) = 1/4`

`x = 1/4`

Solution 3)

`x + 9 = 0`

`x + 9 - color(red)(9) = 0 - color(red)(9)`

`x + 0 = -9`

`x = -9`

The solution is: `x = -1/2` and `x = 1/4` and `x = -9`

Answer 2

`x=-1/2`
`x=1/4`
`x=-9`

Explanation:

If any of paranthesises ( ) will be equal to 0, result will be 0 too.

`2x+1=0`
`2x=-1`
`x=-1/2`

OR

`4x^2+35x-9`
`(4x-1)(x+9)=0`

SO, from THIS we have:

`4x-1=0`
`4x=1`
`x=1/4`

AND

`x+9=0`
`x=-9`

THE ANSWER:

`x=-1/2`
`x=1/4`
`x=-9`