Respuesta :
The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4
Solution:
We have been given a cubic polynomial.
[tex]x^{3}+7 x^{2}+12 x=0[/tex]
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:
[tex]x^{3}+7 x^{2}+12 x=0[/tex]
[tex]x\left(x^{2}+7 x+12\right)=0[/tex] ----- 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:
[tex]x^{2}+7 x+12=0[/tex]
we have to use the quadratic equation to solve this polynomial. The quadratic formula is:
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
Now, a = 1, b = 7 and c = 12
By substituting the values of a,b and c in the quadratic equation we get;
[tex]\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}[/tex]
Therefore, the two roots are:
[tex]\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}[/tex]
And,
[tex]\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}[/tex]
Hence, the three roots of the given cubic polynomial is 0, -3 and -4
