Hey!
The first step to solving this equation would be to factor
![f ( x )](https://tex.z-dn.net/?f=f%20%28%20x%20%29)
. Now since
![f ( x )](https://tex.z-dn.net/?f=f%20%28%20x%20%29)
cannot be factored any further, this will be its final form.
The next step would be to factor the right side of the equation. To do this we'll first break the expression into groups.
<em>Original Equation :</em>
![f ( x ) = x^{2} -10x - 24](https://tex.z-dn.net/?f=f%20%28%20x%20%20%29%20%3D%20%20x%5E%7B2%7D%20-10x%20-%2024%20)
<em>New Equation {Equation Grouped} :</em>
![= ( x^{2} +2x)+(-12x -24)](https://tex.z-dn.net/?f=%3D%20%20%28%20%20x%5E%7B2%7D%20%2B2x%29%2B%28-12x%20-24%29)
Now we must factor out
![x](https://tex.z-dn.net/?f=x)
from
![x^{2} +2x](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%2B2x)
into
![x(x+2)](https://tex.z-dn.net/?f=x%28x%2B2%29)
. We'll also have to factor out -12 from
![-12x-24](https://tex.z-dn.net/?f=-12x-24)
into
![-12(x+2)](https://tex.z-dn.net/?f=-12%28x%2B2%29)
.
<em>Old Equation :</em>
![= ( x^{2} +2x)+(-12x -24)](https://tex.z-dn.net/?f=%3D%20%28%20x%5E%7B2%7D%20%2B2x%29%2B%28-12x%20-24%29)
<em>New Equation :</em>
![=x(x+2)-12(x+2)](https://tex.z-dn.net/?f=%3Dx%28x%2B2%29-12%28x%2B2%29)
Finally, we must factor out the common term that is
![(x+2)](https://tex.z-dn.net/?f=%28x%2B2%29)
.
<em>Old Equation :</em>
![=x(x+2)-12(x+2)](https://tex.z-dn.net/?f=%3Dx%28x%2B2%29-12%28x%2B2%29)
<em>New Equation :</em>
![=(x+2)(x-12)](https://tex.z-dn.net/?f=%3D%28x%2B2%29%28x-12%29)
<em>So, this means that the answer is </em>
.Hope this helps!
- Lindsey Frazier ♥