How do you solve (2x + 1) ( 4x ^ { 2} + 35x - 9) = 0(2x+1)(4x2+35x−9)=0?

2 Answers
smendyka
Mar 16, 2017

See the entire solution process below:

Explanation:

First, factor the binomial to:

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

Now, solve each term for 00:

Solution 1)

2x + 1 = 02x+1=0

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

2x + 0 = -12x+0=−1

2x = -12x=−1

(2x)/color(red)(2) = -1/color(red)(2)2x2=−12

(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

kamil9234
Mar 16, 2017

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

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