When is a polynomial factored completely?

Answer 1

It is factored completely if it can not be factored any more. Quite simply, when all like terms have been combined, and all common factors taken out.

2x + 10y + 8x - original equation

2x + 8x + 10y - rearrange equation

(2x + 8x) + 10y - combine terms

10x + 10y - note common factor

10(x+y) - factored

14xy + 10y + 6 - original equation

2(7xy + 5y + 3) - factor out two.

2(y[7x + 5] + 3) - factor out y. This is factored completely.

10x + 3xy + 6y. It is really hard to make a call on this one. You can factor it either like x(10+3y) + 6y, or 10x + 3y(x + 2).

However, polynomials become more complicated that this, however. Try factoring this:

x2 - 4x + 4

That isn't so easy to factor because there are no like terms. However, it is easy to note that the third term is positive. It is also easy to note that the second term is negative. So, you know that if it is able to be factored, that the factors will both have a minus sign. Using a little deductive reasoning, you'll find that this factors down to (x-2)(x-2), or (x-2)2.

How about this?

x4 - 1

Note that despite the fact that there is an x4 in this, it still follows the rules of difference of squares. Thus can be factored

(x2 - 1)(x2 + 1)

Looking at this, you can see, again, another difference of squares. So, we continue factoring.

(x+1)(x-1)(x2 + 1). Now comes probably to the heart of the question. How do we know that x2 + 1 is factored as far as it can go? x2 + 1 can be rewritten as x2 + 0x + 1. We can note in this instance that the second and third terms are both positive. That means for this to work, there must be case where two positive terms added together equal zero. (x + )(x + ). By logical determination, we can conclude that there is no such number that makes this possible. Thus, x2 + 1 is factored as far as it can go, and thus:

(x+1)(x-1)(x2 + 1)

is also completely factored.

Another example:

x2 -5x - 14

First of all, we start by making observations. Both second and third terms are negative. That means, when we factor this, each binomial factor will have a different sign. We can start by writing out the ground work. (x + )(x - ). Then we try to come up with divisors of 14. In this case: 1, 2, 7, 14. We need to find two that when subtracted equal 5 (different signs equal subtraction when combining like terms). In this case, 7 and 2. Note, for the second term, 5, to be negative, the larger of the two numbers must be negative (-7 + 2 = -5) So, this equation factored is

(x + 2)(x - 7)