# Factoring : use of the distributive property

Factoring : use of the distributive property

Recall that 2 and 3 are said to be factors of 6 because the product of 2 and 3 is 6. Likewise, in an indicated product such as 7ab, the 7, a, and b are called factors of the product. If a positive integer greater than 1 has no factors that are positive integers other than itself and 1, then it is called a prime number. Thus, the prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19. A positive integer greater than 1 that is not a prime number is called a composite number. The composite numbers less than 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, and 18.

Every composite number can be expressed as the product of prime numbers. Consider the following examples.

4 = 2. 2

12 = 2. 2. 3

35 = 5. 7

63 = 3. 3. 7

121 = 11. 11

The indicated product form containing only prime factors is called the prime factorization form of a number. Thus, the prime factorization form of 63 is 3.3.7. we also say that the number has been completely factored when it is expressed in the prime factorization form.

In general, factoring is the reverse of multiplication. Previously, we have used the distributive property to find the product of a monomial and a polynomial as the next examples illustrate.

3 (x + 2) = 3(x) + 3(2)= 3x + 6 ;

5(2x – 1) = 5(2x) – 5(1) = 10x – 5

$x(x^2+6x-4)=x(x^2)+x(6x)-x(4)=x^3+6x^2-4x$

Now, we shall also use the distributive property (except in the form ab + ac = a (b + C) to reverse the process, that is, to factor a given polynomial. Consider the following examples.

3x + 6 = 3(x) + 3(2) = 3 (x + 2)

10x – 5 = 5 (2x) – 5(1) = 5 (2x – 1 )

$x^3+6x^2-4x=x(x^2)+x(6x)-x(4)=x(x^2+6x-4)$

Note that in each example a given polynomial has been factored into the product of a monomial and a polynomial.

Obviously, polynomials could be factored in a variety of ways. Consider some factorizations of $3x^2+12x$.

$3x^2+12x=3x(x+4)\text{ or }3x^2+12x=3(x^2+4x)\text{ or }$

$3x^2+12x=x(3x+12)\text{ or }3x^2+12x=\frac{1}{2}(6x^2+24x)$

We are, however, primarily interested in the first of the above factorization forms and shall refer to it as the completely factored form. A polynomial with integral coefficients is in completely factored form if :

1. It is expressed as a product of polynomials with integral coefecients.
2. No polynomials, other than a monomial, within the factored form can be further factored into polynomials with integral coefficients.

Do you see why only the first of the above factores forms of $3x^2+12x$ is said to be in completely factored form?. In each of the other three forms the polynomial inside the parentheses can be further factored. Furthermore, in the last form, $\frac{1}{2}(6x^2+24x),$ the condition of using only integral coefficients is violated.

The factoring process being discussed in this section is often reffered to as factoring out the highest common monomial factor. The key idea in this process is to recognize the monomial factor which is common to all terms. For example, observe that each term of the polynomial $2x^3+4x^2+6x$ has a factor of 2x. thus, we can write

$2x^3+4x^2+6x=2x\text{( )}$

And insert within the parentheses the appropriate polynomial factor. The terms of this polynomial factor are determined by dividing each term of the original polynomial by the factor 2x. the final completely factored form is

$2x^2+4x^2+6x=2x(x^2+2x+3)$

The following examples further illustrate this process of “factoring out the highest common monomial factor”.

$12x^3+16x^2=4x^2(3x+4)$

$8ab-18b=2b(4a-9)$

$6x^2y^3+27xy^4=3xy^3(2x+9y)$

$30x^3+42x^4-24x^5=6x^3(5+7x04x^2)$

Note that in ech example the common monomial factor itself is not expressed in a completely factored form. For example : $4x^2(3x+4)$ is not written as 2. 2. x . x . (3x + 4).

Sometimes there may be a common binomial factor rather than a common monomial factor. For example, each of the two terms of expression x ( y + 2) + z (y+ 2) has a binomial factor of (y + 2). Thus, we can factor (y + 2) from ech term and our result is as follows;

$x(y+2)+z(y+2)=(y+2)(x+z)$

Consider a few more examples involving a common binomial factor.

$a^2(b+1)+2(b+1)=(b+1)(a^2+2)$

$x(2y-1)-y(2y-1)=(2y-1)(x-y)$

$x(x+2)+3(x+2)=(x+2)(x+3)$

It may be that the original polynomial exhibits no apparent common monomial or binomial factor as is the case with ab + 3a + bc + 3c. However, by factoring a from the first two terms and c from the last two terms it can be expressed as :

$ab+3a+bc+3c=a(b+3)+c(b+3)$

Now a common binomial factor of (b + 3) is obvious and we can proceed as before.

$a(b+3)+c(b+3)=(b+3)(a+c)$

This factoring process is reffered to as factoring by grouping.

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