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 $ax^{2}+bx+c=0$.
$$x^{2}+6x+3=0$$
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.
$$x=\frac{-6±\sqrt{6^{2}-4\times 3}}{2}$$
Square $6$.
$$x=\frac{-6±\sqrt{36-4\times 3}}{2}$$
Multiply $-4$ times $3$.
$$x=\frac{-6±\sqrt{36-12}}{2}$$
Add $36$ to $-12$.
$$x=\frac{-6±\sqrt{24}}{2}$$
Take the square root of $24$.
$$x=\frac{-6±2\sqrt{6}}{2}$$
Now solve the equation $x=\frac{-6±2\sqrt{6}}{2}$ when $±$ is plus. Add $-6$ to $2\sqrt{6}$.
$$x=\frac{2\sqrt{6}-6}{2}$$
Divide $-6+2\sqrt{6}$ by $2$.
$$x=\sqrt{6}-3$$
Now solve the equation $x=\frac{-6±2\sqrt{6}}{2}$ when $±$ is minus. Subtract $2\sqrt{6}$ from $-6$.
$$x=\frac{-2\sqrt{6}-6}{2}$$
Divide $-6-2\sqrt{6}$ by $2$.
$$x=-\sqrt{6}-3$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-3+\sqrt{6}$ for $x_{1}$ and $-3-\sqrt{6}$ for $x_{2}$.
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 $x^2+Bx+C=0$.
$$x ^ 2 +6x +3 = 0$$
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 $rs = C$
$$r + s = -6 $$ $$ rs = 3$$
Two numbers $r$ and $s$ sum up to $-6$ exactly when the average of the two numbers is $\frac{1}{2}*-6 = -3$. 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 $u$.
$$r = -3 - u$$ $$s = -3 + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 3$
$$(-3 - u) (-3 + u) = 3$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$9 - u^2 = 3$$
Simplify the expression by subtracting $9$ on both sides
$$-u^2 = 3-9 = -6$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
$$u^2 = 6$$ $$u = \pm\sqrt{6} = \pm \sqrt{6} $$
The factors $r$ and $s$ are the solutions to the quadratic equation. Substitute the value of $u$ to compute the $r$ and $s$.
