There are several factoring methods used to factor polynomials. A typical approach is to first factor out any common monomial, then to check if the result is one of the special product patterns, and if not, to use some other method (guess and check, "ac" method, etc.).
Three special product patterns you should know for quadratic polynomials are:
(1) The difference of two squares `a^2-b^2=(a+b)(a-b)`
(2) Square of a binomial: `a^2+2ab+b^2=(a+b)(a+b)=(a+b)^2`
(3) Square of a binomial: `a^2-2ab+b^2=(a-b)(a-b)=(a-b)^2`.
One other hint: if the quadratic is in standard form `ax^2+bx+c`, the discriminant is `Delta=b^2-4ac` . In order for the quadratic to factor over the rationals, the discriminant must be a nonnegative square number.
First factor out the common factor of 9 to get `9(16-p^2)`.
The quadratic in the parenthesis is a difference of two squares:
If you didn't see the common factor, you might have recognized the difference of two squares immediately:
`((12)^2-(3p)^2)=(12+3p)(12-3p)`, but each binomial has a common factor of 3, so it is not fully factored.
`3(4+9)3(4-p)=9(4+p)(4-p)`, as before.
A quadratic that factors in the rationals is fully factored if it is written as the product of two linear factors (possibly with a coefficient) where each of the linear factors has no common factor.
Note that there is no common factor, nor is this one of the special product patterns listed above.
We want to write this as a product `(w-p)(w-q)`.
Multiplying, we get `w^2-pw-qw+pg`.
So we need `pq=54, p+q=15` (relating the like variables: `w^2=w^2, -15w=-pw-qw, 54=pq` ).
Running through the list of ways to factor 54, 1*54,2*27,3*18,6*9, we find that 6*9=54 and 6+9=15.
So, the factorization is (w-6)(w-9).
* You can always check to see if the product is the original polynomial. This won't tell you if it is fully factored but will let you know if you made a mistake.
** As noted above, for `w^2-15w+54`, we have `Delta=225-4(1)(54)=9`, which is a perfect square, so we know the polynomial factors.