Answer: (x-1)(x-3)(x+5)
Step-by-step explanation:
Use the rational root theorem
Constant term = 15, Coefficient of leading term = 1
Factors of 15: 1, 3 , 5, 15
Factors of 1: 1
Therefore, (x-1) is 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]
Cubic equations must have 3 factors to get to the third degree, therefore (x-1)(x-3)(x+5)
