Question

How do you solve `3x ^ { 3} + 18= x ( 2x + 27)`?

Answer 1

`x = 3, -3, 2/3`

Explanation:

First multiply out the bracket on the LHS of the equation.

`3x^3 + 18 = 2x^2 + 27x`

Rearrange to get a cubic equation.

`3x^3 + 18 - 2x^2 - 27x = 0`

`3x^3 - 2x^2 - 27x + 18 = 0`

Now find a number that can be substituted into `x` that will make the equation equal zero.

When `x` = 3, the equation is zero.

We now have a factor and now we can factorise the equation.

`(x-3)(3x^2+7x-6) -> (x-3)(3x-2)(x+3)`

Now you can find the answers for `x`.

`x = 3, -3, 2/3`

Answer 2

`x = 3, -3, 2/3`

Explanation:

I would start by distributing the x on the right side of the equal sign.

`3x ^ { 3} + 18= 2x^2 + 27x`

Get everything to one side, I moved everything to the left side. Do this by subtracting the terms on the left side. You get

`3x ^ { 3} + 18 -2x^2 -27x = 0`

Then reorder into standard form (decrease exponents)

`3x^3 -2x^2 -27x + 18 = 0`

When ever you have 4 terms, it is usually a factoring by grouping problem. Put parenthesis around the first two and the last two terms.
`(3x^3 -2x^2) (-27x + 18 )= 0`

Look for what can be factored out of the first grouping: `x^2`
Look for what can be factored out of the second groping: `-9`

The first grouping now looks like `x^2(3x-2)`
The second grouping now looks like `-9(3x-2)`

Notice the terms in the parenthesis are the same.

your problem now looks like:
`x^2 (3x-2) -9(3x -2) = 0`

Now when you factor by grouping you put the two terms in front of the parenthesis together in front of the set of parenthesis.

`(x^2 - 9)(3x-2) = 0`

Now set each sit of parenthesis equal to zero and solve

#(x^2-9) = 0 and (3x-2) =0.

I recognized the first terms as the difference of two squares which means it can be easily factor
`(x^2-9) -> (x+3)(x-3) = 0`

so x could be +3 or -3.

Don't forget to solve the second equation `3x -2 =0`
Add two to both sides, then divide by 3.

`x =2/3`

`x = 3, -3, 2/3`