Since the problem does not specify the polynomial, I suggest you to consider the next polynomial: x^2-7x+6.

You need to form a set comprising all divisors of constant term 6 such that:

D:{-1,-2,-3,-6,1,2,3,6}

You need to find two of these divisors that multipliesd yield 6 such that:

1*6; 2*3,(-1)*(-6),(-2)(-3)

You need to find two of these divisors that added yield 7 such that:

1+6=7

This is the only pair that yields 7.

Hence, since the divisors 1 and 6 respect the simultaneous conditions, the trial factor method lays stress on these two divisors, thus 1 and 6 are the roots of polynomial.

You may write the factored form of polynomial such that:

**x^2-7x+6=(x-1)(x-6)**

This is a very vague question. A polynomial has to be created and its factors determined.

- Let the polynomial be f(x) = x^2 - 4

The factors of this polynomial are (x - 2) and (x + 4)

- Consider the polynomial f(x) = x^2 + 2x + 1

It can be seen that x^2 + 2x + 1 = (x + 1)^2

The factor of f(x)=x^2 + 2x + 1 is (x + 1)

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