Answer:
[tex]8p^2-\frac{12}{5}p^2+\frac{6}{25}p-\frac{1}{125}=\frac{(p+\frac{11}{140})(p+\frac{1}{125})}{125}[/tex]
Step-by-step explanation:
You have the following polynomial:
[tex]8p^2-\frac{12}{5}p^2+\frac{6}{25}p-\frac{1}{125}[/tex] (1)
In order to obtain the factors of the previous polynomial, first you can multiply the polynomial for 125 to obtain only integer coefficients, but also you have to divide for 125, as follow:
[tex](8p^2-\frac{12}{5}p^2+\frac{6}{25}p-\frac{1}{125})\frac{125}{125}\\\\=\frac{1000p^2-300p^2+30p-1}{125}[/tex]
[tex]=\frac{700p^2+30p-1}{125}[/tex] (2)
Next, you use the quadratic formula to get the factors of the numerator of the previous expression:
[tex]p_{1,2}=\frac{-30\pm\sqrt{(30)^2-4(700)(-1)}}{2(700)}\\\\p_{1,2}=\frac{-30\pm 80}{1400}\\\\p_1=-\frac{11}{140}\\\\p_2=-\frac{1}{28}[/tex]
The factors are (p-p1) and (p-p2), then you have in the equation (2) the following factorization:
[tex]8p^2-\frac{12}{5}p^2+\frac{6}{25}p-\frac{1}{125}=\frac{(p+\frac{11}{140})(p+\frac{1}{125})}{125}[/tex]
