Consider $M^{2}-5M+6$. Factor the expression by grouping. First, the expression needs to be rewritten as $M^{2}+aM+bM+6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-5$$ $$ab=1\times 6=6$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $6$.
$$-1,-6$$ $$-2,-3$$
Calculate the sum for each pair.
$$-1-6=-7$$ $$-2-3=-5$$
The solution is the pair that gives sum $-5$.
$$a=-3$$ $$b=-2$$
Rewrite $M^{2}-5M+6$ as $\left(M^{2}-3M\right)+\left(-2M+6\right)$.
$$\left(M^{2}-3M\right)+\left(-2M+6\right)$$
Factor out $M$ in the first and $-2$ in the second group.
$$M\left(M-3\right)-2\left(M-3\right)$$
Factor out common term $M-3$ by using distributive property.
$$\left(M-3\right)\left(M-2\right)$$
Rewrite the complete factored expression.
$$3\left(M-3\right)\left(M-2\right)$$
Steps Using the Quadratic Formula
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 $ax^{2}+bx+c=0$.
$$3M^{2}-15M+18=0$$
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.
