Factor the expression by grouping. First, the expression needs to be rewritten as $a^{2}+pa+qa+2$. To find $p$ and $q$, set up a system to be solved.
$$p+q=-3$$ $$pq=1\times 2=2$$
Since $pq$ is positive, $p$ and $q$ have the same sign. Since $p+q$ is negative, $p$ and $q$ are both negative. The only such pair is the system solution.
$$p=-2$$ $$q=-1$$
Rewrite $a^{2}-3a+2$ as $\left(a^{2}-2a\right)+\left(-a+2\right)$.
$$\left(a^{2}-2a\right)+\left(-a+2\right)$$
Factor out $a$ in the first and $-1$ in the second group.
$$a\left(a-2\right)-\left(a-2\right)$$
Factor out common term $a-2$ by using distributive property.
$$\left(a-2\right)\left(a-1\right)$$
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 $ax^{2}+bx+c=0$.
$$a^{2}-3a+2=0$$
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.
Now solve the equation $a=\frac{3±1}{2}$ when $±$ is plus. Add $3$ to $1$.
$$a=\frac{4}{2}$$
Divide $4$ by $2$.
$$a=2$$
Now solve the equation $a=\frac{3±1}{2}$ when $±$ is minus. Subtract $1$ from $3$.
$$a=\frac{2}{2}$$
Divide $2$ by $2$.
$$a=1$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $2$ for $x_{1}$ and $1$ for $x_{2}$.
$$a^{2}-3a+2=\left(a-2\right)\left(a-1\right)$$
Steps Using Direct Factoring Method
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.
$$x ^ 2 -3x +2 = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = 3 $$ $$ rs = 2$$
Two numbers $r$ and $s$ sum up to $3$ exactly when the average of the two numbers is $\frac{1}{2}*3 = \frac{3}{2}$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
$$r = \frac{3}{2} - u$$ $$s = \frac{3}{2} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 2$
$$(\frac{3}{2} - u) (\frac{3}{2} + u) = 2$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{9}{4} - u^2 = 2$$
Simplify the expression by subtracting $\frac{9}{4}$ on both sides
$$-u^2 = 2-\frac{9}{4} = -\frac{1}{4}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
