We want to rewrite the general quadratic equation \(ax^2 + bx + c\) in the form \((\alpha x + \beta)(\gamma x + \delta)\).
To do this we look for two numbers \(p\) and \(q\) such that \(ac = pq\) and \(p + q = b\). By convention we choose \(p \leq q\).
We can now rewrite the general quadratic equation as …
\[ax^2 + (p + q)x + c\] \[= ax^2 + px + qx + c\] \[= px(\frac{a}{p}x + 1) + c(\frac{q}{c}x + 1)\]Now, since \(ac = pq\) then \(\frac{a}{p} = \frac{q}{c}\).
Therefore we can further write the general quadratic equation as …
\[px(\frac{a}{p}x + 1) + c(\frac{a}{p}x + 1)\]or …
\[px(\frac{q}{c}x + 1) + c(\frac{q}{c}x + 1)\]We could do either but let’s use \(\frac{a}{p}\) since we chose \(p\) to be the smaller number.
So …
\[px(\frac{a}{p}x + 1) + c(\frac{a}{p}x + 1)\]We now have a common factor of \((\frac{a}{p}x + 1)\) so let’s pull that out …
\[= (\frac{a}{p}x + 1)(px + c)\]And rewrite as …
\[= \frac{(ax + p)(px + c)}{p}\]And thus we have a formula to factor the general quadratic once we have found \(p\) and \(q\) such that \(ac = pq\) and \(p + q = b\). We could use either \(p\) or \(q\) but, by convention, we use the smaller of the two and call it \(p\).
\[ax^2 + bx + c = \frac{(ax + p)(px + c)}{p}\]Let’s try some examples.
Example 1
\[6x^2 + 11x + 3\]\(a = 6\), \(b = 11\) and \(c = 3\); \(ac = 6 \cdot 3 = 18\).
\(18 = 1 \cdot 18 = 2 \cdot 9 = 3 \cdot 6\). Of these pairs, \(2 + 9 = 11\) and so we can take \(p = 2\).
Therefore …
\[6x^2 + 11x + 3 = \frac{(6x + 2)(2x + 3)}{2} = (3x + 1)(2x + 3)\]Example 2
\[6x^2 + 13x + 6\]\(a = 6\), \(b = 13\) and \(c = 6\); \(ac = 6 \cdot 6 = 36\).
\(36 = 1 \cdot 36 = 2 \cdot 18 = 4 \cdot 9\). Of these pairs, \(4 + 9 = 13\) and so we can take \(p = 4\).
Therefore …
\[6x^2 + 13x + 6 = \frac{(6x + 4)(4x + 6)}{4} = \frac{(6x + 4)(4x + 6)}{2 \cdot 2} = \frac{(6x + 4)}{2}\frac{(4x + 6)}{2} = (3x + 2)(2x + 3)\]Example 3
\[9x^2 + 12x - 12\]\(a = 9\), \(b = 12\) and \(c = -12\); \(ac = 9 \cdot -12 = -108\).
\(-108 = -1 \cdot 108 = -2 \cdot 54 = -4 \cdot 26 = -6 \cdot 18 = -8 \cdot 13\). of these pairs, \(-6 + 18 = 12\) and so we can take \(p = -6\).
Therefore …
\[9x^2 + 12x - 12 = \frac{(9x - 6)(-6x - 12)}{-6} = \frac{(9x - 6)(-6x - 12)}{3 \cdot -2} = \frac{(9x - 6)}{3}\frac{(-6x - 12)}{-2} = (3x - 2)(3x + 6)\]