Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$3b^{2}+9b=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$b=\frac{-9±\sqrt{9^{2}}}{2\times 3}$$
Take the square root of $9^{2}$.
$$b=\frac{-9±9}{2\times 3}$$
Multiply $2$ times $3$.
$$b=\frac{-9±9}{6}$$
Now solve the equation $b=\frac{-9±9}{6}$ when $±$ is plus. Add $-9$ to $9$.
$$b=\frac{0}{6}$$
Divide $0$ by $6$.
$$b=0$$
Now solve the equation $b=\frac{-9±9}{6}$ when $±$ is minus. Subtract $9$ from $-9$.
$$b=-\frac{18}{6}$$
Divide $-18$ by $6$.
$$b=-3$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $0$ for $x_{1}$ and $-3$ for $x_{2}$.
$$3b^{2}+9b=3b\left(b-\left(-3\right)\right)$$
Simplify all the expressions of the form $p-\left(-q\right)$ to $p+q$.
