Since this is a quadratic polynomial of the form:
[tex]P(x)=ax^2+bx+c[/tex]
We can use the quasratic formula to get the factors of the polynomial.
The quadratic formula is:
[tex]x_{1,2}=\frac{-b\pm\sqrt[\square]{b^2-4\cdot a\cdot c}}{2\cdot a}[/tex]
In this case, a = 6, b = -5, c = -6
Then:
[tex]\begin{gathered} x_{1,2}=\frac{-(-5)\pm\sqrt[\square]{(-5)^2-4\cdot6\cdot(-6)}}{2\cdot6} \\ \end{gathered}[/tex]
Then solve:
[tex]\begin{gathered} x_{1,2}=\frac{5\pm\sqrt[\square]{25^{}-(-144)}}{12} \\ x_{1,2}=\frac{5\pm\sqrt[\square]{169}}{12} \\ x_{1,2}=\frac{5\pm13}{12} \end{gathered}[/tex]
Now we can find the two roots:
[tex]\begin{gathered} x_1=\frac{5+13}{12}=\frac{18}{12}=\frac{3}{2} \\ x_2=\frac{5-13}{12}=-\frac{8}{12}=-\frac{2}{3} \end{gathered}[/tex]
Then the factored form of the polynomial is:
[tex]P(x)=(x-\frac{3}{2})(x+\frac{2}{3})[/tex]
With p as the variable:
[tex]6p-5p-6p=(p-\frac{3}{2})(p+\frac{2}{3})[/tex]
