Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-4$$ $$ab=1\times 3=3$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. The only such pair is the system solution.
$$a=-3$$ $$b=-1$$
Rewrite $x^{2}-4x+3$ as $\left(x^{2}-3x\right)+\left(-x+3\right)$.
$$\left(x^{2}-3x\right)+\left(-x+3\right)$$
Factor out $x$ in the first and $-1$ in the second group.
$$x\left(x-3\right)-\left(x-3\right)$$
Factor out common term $x-3$ by using distributive property.
$$\left(x-3\right)\left(x-1\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}-4x+3=0$$
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 $x=\frac{4±2}{2}$ when $±$ is plus. Add $4$ to $2$.
$$x=\frac{6}{2}$$
Divide $6$ by $2$.
$$x=3$$
Now solve the equation $x=\frac{4±2}{2}$ when $±$ is minus. Subtract $2$ from $4$.
$$x=\frac{2}{2}$$
Divide $2$ by $2$.
$$x=1$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $3$ for $x_{1}$ and $1$ for $x_{2}$.
$$x^{2}-4x+3=\left(x-3\right)\left(x-1\right)$$
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 $x^2+Bx+C=0$.
$$x ^ 2 -4x +3 = 0$$
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 $rs = C$
$$r + s = 4 $$ $$ rs = 3$$
Two numbers $r$ and $s$ sum up to $4$ exactly when the average of the two numbers is $\frac{1}{2}*4 = 2$. 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 $u$.
$$r = 2 - u$$ $$s = 2 + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 3$
$$(2 - u) (2 + u) = 3$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$4 - u^2 = 3$$
Simplify the expression by subtracting $4$ on both sides
$$-u^2 = 3-4 = -1$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
$$u^2 = 1$$ $$u = \pm\sqrt{1} = \pm 1 $$
The factors $r$ and $s$ are the solutions to the quadratic equation. Substitute the value of $u$ to compute the $r$ and $s$.
