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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)

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Related Questions

What is the polynomial when each of its factors is prime?

Completely Factored


How do you know if a polynomial is in factored form?

You can't know if a general polynomial is in factored form.


A polynomial that can be factored is called?

It is still called a polynomial.


Factor this polynomial completely a-ab-42b?

Too bad that's not a^2 - ab - 42b^2 That factors to (a + 6b)(a - 7b)


A polynomial that can not be factored is called?

If a number cannot be factored it is a prime number.


What is the value of having factored form of a polynomial?

The factored form of a polynomial is valuable because it simplifies the process of finding its roots or zeros, making it easier to solve equations. It also provides insights into the polynomial's behavior, such as identifying multiplicities of roots and understanding its graph. Additionally, factored form can facilitate polynomial division and help in applications such as optimization and modeling in various fields.


Why can't a polynomial be factored?

It can, so the question does not make sense.


What is the factored form of the polynomial x2 plus 9x plus 20?

It is (x+4)(x+5) when factored


What is 27-y3 factored completely?

27-y3 factored completely = 24


How is 15x3 plus 20x factored as a polynomial?

5x(3x+4)


When is an expression factored completely?

When the expression is broken down into its prime factors it is factored completely.


What is a prime polynomial?

A prime polynomial is a polynomial that cannot be factored into the product of two non-constant polynomials over its coefficient field. In other words, it has no divisors other than itself and the unit (constant) polynomials. For example, in the field of real numbers, (x^2 + 1) is a prime polynomial because it cannot be factored into real linear factors. Conversely, polynomials like (x^2 - 1) are not prime because they can be factored as ((x - 1)(x + 1)).