Q:

# ​Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Accepted Solution

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