The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4
<u>Solution:</u>
We have been given a cubic polynomial.

We need to find the three roots of the given polynomial.
Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.
This gives us:

----- eqn 1
So, from the above eq1 we can find the first root of the polynomial, which will be:
x = 0
Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

Now, a = 1, b = 7 and c = 12
By substituting the values of a,b and c in the quadratic equation we get;

<em><u>Therefore, the two roots are:</u></em>

And,

Hence, the three roots of the given cubic polynomial is 0, -3 and -4