# Solve 3x^3 - 21x^2 + 36 x =0

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To solve 3x^3-21x^2+36x = 0

We factor the left.

3x(x^2-7x+12) = 0............(1)

We take x^2-7x+12

We split the middle tern -7x = -4x and -3x such that their product is equal to the product o first and last term , that is, x^2 and 12.

x^2-7x+12

x^2-7x+12=x^2-4x-3x+12

x^2-7x+12=x(x-4)-3(x-4)

x^2-7x+12=(x-4)(x-3)

Therefore 3x^3-21x^2+36x = 0 could be rewritten as:

3x(x-4)(x-3) = 0.

Therefore by zero product rule we get:

x = 0 , x-4 = 0 , x -3 = 0.

x = 0 , or x = 4 , or x = 3.

We have to solve 3x^3 - 21x^2 + 36 x =0 for x.

We see that the highest power of x in this polynomial is 3. So we are looking for three roots.

First we’ll extract x and any other common factors that all the terms have.

3x^3 - 21x^2 + 36 x =0

=> 3x [x^2 – 7x^2 + 12] =0

Now let’s write 7x as 3x + 4x

=> 3x [x^2 – 4x – 3x +12] =0

=> 3x[x (x - 4) – 3 (x -4)] =0

=> 3x (x-3) (x-4) =0

Now we have the three roots: x = 0, x = 3 and x = 4

**Therefore the required roots of the equation are x= 0, x= 3 and x =4**