How do you solve x2−6x−16=0 by factoring?

Question

Maritza Mccall · Accepted Answer

Factoring a polynomial p(x) means to find its roots

x_{i}

, and divide the polynomial by the term

(x - x_{i})

, writing p(x) as

(x - x_{i}) q (x)

, and then iterating the process searching the roots of q(x) (which of course are roots of p(x), too. The fundamental theorem of algebra states that you can always write a polynomial of degree k as the product of k linear factors

(x - x_{i})

, i.e., every polynomial of degree k has k complex roots.
These roots can be complex but not real, and in that case, you cannot simplify the polynomial in R.
In your case, the solutions are easy to find, without doing any calculation: you can remember the rule which states that, if you have a quadratic of the form

x^{2} - s x + p

, the product of the solutions equals p and the sum of the solutions equals s.
So, we have to find two numbers which sum up to -6, and that multiplied one for the other give -16. You can easily convince yourself that these two numbers are -2 and 8.
If -2 and 8 are roots of our polynomial, for what we said above the following must hold:

x^{2} - 6 x - 16 = (x + 2) (x - 8)

How do you solve x^{2}-6x-16=0 by factoring?

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