Answer: (x-3)(x-1)(x+5)
Step-by-step explanation:
Use the rational root theorem
A0 = 15, An = 1
Factors of 15: 1, 3 , 5, 15
Factors of 1: 1
Using (x-1) as a factor
[tex]\begin{array}{l}=\frac{\left(x-1\right)\frac{x^3+x^2-17x+15}{x-1}}{x-3}\\\\=\frac{\left(x-1\right)\left(x^2+2x-15\right)}{\left(x-3\right)}\\\\=\frac{\left(x-1\right)\left(x-3\right)\left(x+5\right)}{x-3}\\\\=\left(x-1\right)\left(x+5\right)\end{array}[/tex]
A cubic equation cannot have 2 factors, it must have at least 3 to get ti the 3rd degree, so the answer is (x-3)(x-1)(x+5)