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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:
You need to find two of these divisors that multipliesd yield 6 such that:
You need to find two of these divisors that added yield 7 such that:
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:
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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