Another technique for factoring a trinomial of the form |ax^2+bx+c| is by applying the quadratic formula: |\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}.|
A trinomial can be factored with this method if and only if the value of its discriminant, i.e., |b^2-4ac,| is greater than or equal to zero.
If a trinomial of the form |\color{#333fb1}{a}x^2+bx+c| can be factored, then it can also be written in factored form |\color{#333fb1}{a}(x-x_1)(x-x_2)| where |x_1| and |x_2| are the two zeros calculated using the quadratic formula.
Note: If only one value is obtained using the formula, then |x_1=x_2.|
Consider the trinomial |2x^2+3x-1.|
In the trinomial, |a=2,| |b=3| and |c=-1.| To determine the values of |x_1| and |x_2,| it is necessary to apply the quadratic formula. ||\begin{align} x_{1,2} &= \dfrac{-3 \pm \sqrt{(3)^2 - 4(2)(-1)}}{2 (2)} \\ x_{1,2} &= \dfrac{-3 \pm \sqrt{9 -{^{_{\Large{-}}}} 8}}{4} \\ x_{1,2} &= \dfrac{-3 \pm \sqrt{17}}{4} \end{align}||
At this stage, the formula must be separated into two parts due to the |\pm.| ||\begin{align} x_1 &= \dfrac{-3 + \sqrt{17}}{4} \approx 0.28 \\ x_2 &= \frac{-3 - \sqrt{17}}{4} \approx -1.78 \end{align}||
Therefore, we obtain this factorization. ||2x^2+3x-1 = 2(x-0.28)(x+1.78)||
For more accurate results, it is better to use the radicals (roots) directly. ||2x^2+3x-1 = 2 \left(x-\dfrac{-3 + \sqrt{17}}{4}\right)\left(x - \dfrac{-3 - \sqrt{17}}{4}\right)||
Consider the trinomial |x^2+5x+6.|
In the trinomial, |a=1,| |b=5| and |c=6.| To determine the values of |x_1| and |x_2,| it is necessary to apply the quadratic formula.
||\begin{align} x_{1,2} &= \dfrac{-5 \pm \sqrt{(5)^2 - 4 (1)(6)}}{2(1)} \\ x_{1,2} &= \dfrac{-5 \pm \sqrt{25 - 24}}{2} \\ x_{1,2} &= \dfrac{-5 \pm \sqrt{1}}{2} \\ x_{1,2} &= \dfrac{-5 \pm 1}{2} \end{align}||
At this stage, the formula must be separated into two parts due to the |\pm.| ||\begin{align} x_1 &= \dfrac{-5 + 1}{2} = -2 \\ x_2 &= \frac{-5 - 1}{2}= -3 \end{align}||
Therefore, we obtain this factorization. ||\begin{align} x^2+5x+6 &= 1\big(x-(-2)\big)\big(x-(-3)\big) \\ &= 1(x+2)(x+3) \end{align}||
It is not necessary to write the factor |1| in front of the brackets. We can simply write |(x+2)(x+3).|