book says (x+5)(x-3)(x-1)
Help! I do not know procedure to factor trinomial??????????
Let $\displaystyle f(x)=x^3+x^2-17x+15.$
Try a number $\displaystyle a$ and see if $\displaystyle f(a)=0,$ say $\displaystyle a=1.$
$\displaystyle f(1) = 1^3+1^2-17(1)+15 = 0.$
Thus $\displaystyle x-1$ is a factor, i.e. $\displaystyle x^3+x^2-17x+15=(x-1)g(x)$ where $\displaystyle g(x)$ is a quadratic polynomial. Once you've found $\displaystyle g(x)$ 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, $\displaystyle x^2- 3$ can be written as $\displaystyle (x- \sqrt{3})(x+ \sqrt{3})$ 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, $\displaystyle a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$, are of the form $\displaystyle \frac{m}{n}$ with m an integer that evenly divides $\displaystyle a_0$ and n an integer that evenly divides $\displaystyle a_n$. Here, $\displaystyle a_n= 1$ so any rational root must have denominator 1- in other words be an integer. $\displaystyle a_0= 15$ 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.
$\displaystyle (3)^3+ (3)^2- 17(3)+ 15= 27+ 9- 51+ 15= 0$ so 3 is a root and x- 3 is a factor.
$\displaystyle (-3)^3+ (-3)^2- 17(-3)+ 15= -27+ 9+ 51+ 15= 48$ so -3 is not a root and x-(-3)= x+ 3 is not a factor.
$\displaystyle (5)^3+ (5)^2- 17(5)+ 15= 125+ 25- 85+ 15= 80$ so 5 is not a root and x- 5 is not a factor.
$\displaystyle (-5)^3+ (-5)^2- 17(-5)+ 15= -125+ 25+ 85+ 15= 0$ 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
$\displaystyle x^3+ x^2- 17x+ 15= (x+5)(x-3)(x-1)$
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 $\displaystyle x^3+ x^2- 17x+ 15$ by x- 1, leaving $\displaystyle x^2+ 2x- 15$ which is relatively easy to as (x- 3)(x+ 5).