A quadratic equation can be rewritten as
![ax^2+bx+c=a(x-x_1)(x-x_2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/dssedkcg4c691kz7ff1pn69fdovn9pno27.png)
where x1 and x2 represents the roots of the quadratic equation.
To find those roots we can use the quadratic equation. Given a quadratic equation with the following form
![ax^2+bx+c=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/mvkhuzwnjhb4epaf7jjcoq2vi4zdi4350m.png)
its roots are given by
![x_(\pm)=(-b\pm√(b^2-4ac))/(2a)](https://img.qammunity.org/2023/formulas/mathematics/college/nruhb6lm9kr18bxsnvk8bd8e8zzaefm1da.png)
In our first quadratic equation, we have
![x^2+5x-14](https://img.qammunity.org/2023/formulas/mathematics/college/oqr7y9r15b6v8er307bf1quaj4taqt4bui.png)
therefore, its roots are
![\begin{gathered} x_(\pm)=(-(5)\pm√((5)^2-4(1)(-14)))/(2(1)) \\ =(-5\pm√(25+56))/(2) \\ =(-5\pm9)/(2) \\ \implies\begin{cases}x_-={-7} \\ x_+={2}\end{cases} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8rxofbteqysyhwv9qpu2i2kd4jj4058r40.png)
Then, this quadratic expression can be factorized as
![x^2+5x-14=(x+7)(x-2)](https://img.qammunity.org/2023/formulas/mathematics/college/zu40v2pee719vds66o5bignbjsxnk54zy1.png)
Using the same process for the other expression, we have
![x^2+7x+6=(x+1)(x+6)](https://img.qammunity.org/2023/formulas/mathematics/college/sqaz7zmobk1fgb1tzi45t6ohq9vjwusbn7.png)