Let
Try a number and see if say
Thus is a factor, i.e. where is a quadratic polynomial. Once you've found you can factorize it further.
To "factor" a polynomial with integer coefficients typically means "write as a product of factors with integer coefficients. For example, can be written as but we don't normally think of that as "factoring".
What Sylvia104 did was look for rational zeros of the polynomial. There is a "rational root theorem" that says "Any rational number roots of the polynomial equation, , are of the form with m an integer that evenly divides and n an integer that evenly divides . Here, so any rational root must have denominator 1- in other words be an integer. so the only possible rational roots are 1, -1, 3, -3, 5, -5, 15, and -15, the factors of 15.
Check each one: [tex](1)^3+ (1)^2- 17(1)+ 15= 1+ 1- 17+ 15= 0. Yes! 1 is a root so x- 1 is a factor.
[tex](-1)^3+ (-1)^2- 17(-1)+ 15= -1+ 1+ 17+ 15= 32, not 0. -1 is not a root so x-(-1)= x+ 1 is not a factor.
so 3 is a root and x- 3 is a factor.
so -3 is not a root and x-(-3)= x+ 3 is not a factor.
so 5 is not a root and x- 5 is not a factor.
so -5 is a root and x- (-5)= x+ 5 is a factor.
Having found three factors, we don't need to check 15 and -15 as roots- we now know that
I suspect that Sylvia104's point was that having seen that 1+ 1- 17+ 15= 0 so that x= 1 is a zero of the polynomial and x- 1 a factor, you could divide by x- 1, leaving which is relatively easy to as (x- 3)(x+ 5).