# Solve the following equation for x: 3x^3 + 22x^2 - 16x = 0

*print*Print*list*Cite

### 4 Answers

We have to solve the following equation for x, 3x^3 + 22x^2 - 16x =0

We know factorize 3x^3 + 22x^2 - 16x =0

=> x(3x^2 + 22x - 16) = 0

=> x( 3x^2 + 24x - 2x - 16) =0

=> x( 3x( x + 8) - 2(x + 8)) = 0

=> x(3x - 2)(x + 8) = 0

So the first root is x = 0

for 3x - 2 = 0 , we have x = 2/3

for x + 8 = 0, we have x = -8

**Therefore the three roots of x are 0, 2/3 and -8.**

3x^3 + 22x^2 - 16x = 0

First we will factor x from all terms.

==> x*(3x^2 + 22x - 16) = 0

Now we will find the roots for the quadratic equation.

==> x1= (-22 + sqrt(484+4*3*16) / 2*3

= (-22 + 26)/6 = ( 4/6 = 2/3

==> x1= 2/3 ==> **(3x-2)** is a factor for the quadratic equation.

==> x2= ( -22-26)/6 = -48/6 = -8.

==> x2= -8. Then, **(x+8)** is a factor of the quadratic equation.

==> 3x^3 + 22x^2 - 16x = x*(3x^2 + 22x -16)

= x*(3x-2)(x+8)

Then, there are three solutions to the equation.

x1= 0 , x2= 2/3, and x3= -8.

**==> x= { -8, 0, 2/3}.**

x(3x^2 + 22x - 16) = 0

3x^2 + 24x - 2x - 16

(3x^2 + 24x) (- 2x - 16)

3x (x + 8) -2 ( x + 8)

(3x - 2) (x + 8)

x (3x - 2) (x + 8)

x = 0

3x - 2 =0

3x = 2

x = 2/3

x + 8 = 0

x = -8

3x^3 + 22x^2 - 16x = 0

factor out x

x(3x^2 + 22x - 16) = 0

a b c

multiply a by x 3 x -16 = -48

find factors of -48 that subtract to equal b which you be (24 and -2)

Now replace those numbers as b

3x^2 + 24x - 2x - 16

group

(3x^2 + 24x) (- 2x - 16)

factor out the greatest common factors:

3x (x + 8) -2 ( x + 8)

group

(3x - 2) (x + 8)

don't forget that x we factored out first:

**x (3x - 2) (x + 8)**

this is the answer or you can set them equal to 0 and get the solutions

**x = 0**

3x - 2 =0

3x = 2

**x = 2/3**

x + 8 = 0

**x = -8**