Solve for x if 3x^3 + 27x^2 = 30x.

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Given: 3x^3 + 27x^2 = 30x

We first have to put all the terms on one side so the equation becomes,

3x^3 + 27x^2 - 30x = 0

Now, we do common factor

3x( x^2 + 9x - 10) = 0

Then, factor the 2nd term

3x(x + 10)(x - 1) = 0.

Equate the 3 factors to 0,

3x = 0

**x = 0**

x + 10 = 0

**x = -10**

x - 1 = 0

**x = 1**

**x = -10, 0, 1**

**Do not forget to check if the roots satisfy the original equation. **

Given the equality: 3x^3 + 27x^2 = 30x.

We need to solve for x value that satisfies the equality.

First, let us simplify the equality.

We will subtract 30x from both sides.

==> 3x^3 + 27x^2 - 30x = 0

Now we notice that (3x) is a common factor for all terms.

Then, we will factor (3x).

==> 3x*(x^2 + 9x - 10 ) = 0

Now we will factor between the brackets.

==> 3x ( x+10) ( x-1) = 0

**Then we have 3 possible values for x that satisfies the equality.**

**==> x1= 0**

**==> x2= -10**

**==> x3= 1**

**==> x = { 0, 1, -10 }**

We have to find x given that 3x^3 + 27x^2 = 30x.

Now 3x^3 + 27x^2 = 30x

cancel 3

=> x^3 + 9x^2 = 10x

=> x^3 + 9x^2 - 10x = 0

=> x( x^2 + 9x - 10) = 0

=> x( x^2 + 10x - x - 10) = 0

=> x[ x(x +10) - 1(x +10)] =0

=> x(x - 1)(x + 10) = 0

For x - 1 = 0, x = 1

For x + 10 =0 , x = -10

And x = 0

**Therefore x can be 0, 1 and -10.**

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