 ### Factoring

When an algebraic expression is the product of two or more quantities, each of these quantities is called a factor of the expression; and the determination of these factors is called the factoring of the expression.

### Common Monomial Factors

We have already learned that to multiply a multinomial by a monomial we apply the Distributive Law for Multiplication. Algebraically, this rule may be expressed as follows:

a(x-y + z) = ax-ay + az.

Conversely, if we have an expression such as ax-ay+az, it may be written in the factored form a(x-y+z).

In general, if every term of an algebraic expression contains a common factor, the expression may be represented as the product of that common factor and the quotient obtained by dividing the original expression by the common factor.

Example 1 Factor the expression 4x-6

Solution: Each of the two terms contains the common factor 2, hence,

4x-6=2(2x-3).

Example 2 Factor the expression x^3 -2x^2+5x.

Solution: Each of the three terms contains the common factor x, hence,

x^3 -2x^2+5x=x(x^2-2x+5)

Note. In factoring exercises the student should always verify his results, by mentally forming the product of the factors he has chosen.

Example 3 Factor the expression 3a^4b-3a^3b^2+ 6a^2b^2.

Solution: Each of the three terms contains the common factor 3a^2b, hence,

3a^4b-3a^3b^2+ 6a^2b^2=3a^2b(a^2-ab+2b).

Note. In seeking the factor which is common to all terms, one should remember that he is seeking the greatest factor which is common to all terms. In other words, when the original expression is divided by the common factor, the terms of the quotient obtained should not contain any other common factor.

Example 4 Factor the expression a(x + y) + b(x + y).

Solution: Thinking of (x + y) as a single quantity, say p, we may factor as follows:

a(x + y) + b(x + y) = ap + bp = p(a+ b) = (x + y)(a + b).

Note. The above substitution was made only to indicate that the expression (x + y) is considered as a single quantity. When one can factor without actually making such a substitution, do so, as it saves time.

### Common Binomial Factors

Although all of the terms of an expression may not contain a common factor, it is sometimes possible to group the terms so that each group will have a common factor.

Example 1 Factor ac + bc + ad + bd.

Solution: Grouping together the first two terms and the last two terms, we have

(ac + bc) + (ad + bd).

Factoring the expressions within each parentheses, we have

c(a + b) + d(a +b).

We see here that the quantity (a+b) is common to both terms, hence we have

(a + b)(c + d).

Example 2 Factor x^2- ax + bx- ab.

Solution: Following the procedure described in Example 1,

x^2- ax + bx- ab = (x^2-ax) + (bx-ab)

= x(x- a) + b(x-a)

= (x-a)(x + b).

Example 3 Factor 12a^2-4ab-3 ac + bc.

Solution:

12a^2-4ab-3 ac + bc=(12a^2-4ab)-(3ac-bc)

= 4a(3a-b) -c(3a-b)

= (3a-b)(4a-c)

Note. In the first line of work it is usually sufficient to see that each pair contains some common factor. Thus, in Example 3, by a different grouping,

12a^2-4ab-3 ac + bc=(12a^2-3ac)-(4ab-bc)

= 3a(4a-c)-b(4a-c)

= (4a-c)(3a-b)

This is the same result as before, since the order of factors is immaterial.

### Factoring of Simple Trinomials

Before considering the factoring of trinomials, let us recall the manner in which the product of two binomials gives rise to a trinomial. Thus, by earlier lesson, we have:

(x+5)(x+2)=x^2+7x+10

(x-5)(x-2) =x^2-7x+10

(x+5)(x-2) =x^2+3x-10

(x-5)(x+2) =x^2-3x-10

In each of the above products:

1. The first term of the trinomial is the product of the first terms of the binomials.

2. The second term of the trinomial is obtained by finding the algebraic sum of the product of the two outer terms and the two inner terms.

3. The third term of the trinomial is the product of the second terms of the binomials.

We now propose to consider the converse problem. That is, to resolve trinomial expressions, such as x^2+ 2x-15, into their component binomial factors. We observe from the above results that the binomial factors must satisfy the three requirements:

1. The first terms must be factors of x^2.

2. The second terms must be factors of -15.

3. These factors must be chosen in such a manner that when the product of the outer terms of the binomials is combined with the product of the inner terms, the result will be +2x.

The totality of binomial factors which satisfy the first two conditions are:

(x+15)(x-1), (x-15)(x+ 1), (x-5)(x+ 3), (x +5)(x-3).

However, only in the last of these trial factors do we obtain +2x, the required middle term. Hence,

x^2+ 2x-15=(x+5)(x-3).

Example 1 Factor a^2 + 7a + 12.

Solution: The second terms of the factors must be such that their product is +12 and their sum is +7. It is clear that they must be +3 and +4; hence

a^2 + 7a + 12 = (a+ 3)(a+4).

Example 2 Factor x^2-x- 6.

Solution: The second terms of the factors must be such that their product is -6, and their algebraic sum is -1. Hence, they must have opposite signs, and the greater of them must be negative in order to give its sign to their sum. Hence, the required factors are -3 and +2.

x^2-x- 6 = (x-3)(x+ 2).

When the third term of the trinomial is negative, if preferred, the following method may be adopted.

Example 3 Factor x^4+5x^2-104.

Solution: Find two numbers whose product is 104, and whose difference is 5. These are 13 and 8; hence, inserting signs so that the positive may predominate:

x^4+5x^2-104 = (x^2 + 13)(x^2-8).

Note. Always verify results by forming the product mentally.

### Factoring of General Trinomials

When the coefficient of the term of the trinomial having the highest power is not unity, the number of possible trial factors is considerably greater.

It will be helpful in particular to note that

1. If the third term of the trinomial is positive, then the second term of its factors both have the same sign, and this sign is the same as that of the middle term of the trinomial.

2. If the third term of the trinomial is negative, then the second terms of its factors have opposite signs.

Example 1 Factor 5x^2+7x-6.

Solution: Write (5x   2)(x   3) for a first trial. Since 2 and 3 must have opposite signs, the difference between the inner and the outer products must be 7x. However, since (5 × 3)  (2 × 1) = 13 this combination fails to give the correct coefficient of the middle term.

Next try (5x   3)(x   2). Since (5 × 2) (3 × 1) = 7, these factors will be correct if we insert signs so that the positive will predominate.

Thus,       5x^2+7x-6= (5x-3)(x + 2).

Example 2 Factor 7x^2-19x+ 10.

Solution: We note that the factors which give 10 are both negative, and so we may write (7x   )(x ). We now must place factors of 10 in the parentheses, so that the sum of the inner and outer products yields 19. The possible factors of 10 are 10 and 1, or 5 and 2; and since (7 × 2) + (5 × 1) = 19:

7x^2-19x+ 10 = (7x-5)(x-2).

Example 3 Factor 4x^2-12xy + 9y^2.

Solution:

4x^2-12xy + 9y^2=(2x-3y)(2x-3y)=(2x-3y)^2

Example 4 Factor 8 + 6x-5x^2.

Solution:

8 + 6x-5x^2=(4+5x)(2-x).

### Difference of Two Squares

Multiplying (a + b) by (a-b), we obtain the identity:

(a+ b)(a-b)=a^2-b^2

a result which may be stated as follows:

The product of the sum and the difference of any two quantities is equal to the difference of their squares.

Conversely, we also have:

The difference of the squares of any two quantities is equal to the product of their sum and their difference.

Example 1 Factor 9x^2-16y^2.

Solution: 9x^2-16y^2 = (3x)^2-(4y)^2.

Hence, the first factor is the sum of 3x and 4y, and the second factor is the difference of 3x and 4y; thus,

9x^2-16y^2 = (3x+4y)(3x-4y).

Example 2 Factor 1-36m^6.

Solution:     1-36m^6=(1)^2-(6m^3)^2,

= (1 + 6m^3)(1- 6m^3)

Note. The intermediate step may be omitted when the principle is understood.

### Difference of Two Squares When One or Both Are Binomials

When one or both of the squares is a binomial (or, in fact, any multiple expression), we employ the same method as described in the preceding article.

Example 1 Factor (a + 2b)^2-9x^2

Solution: The sum of a+ 2b and 3x is a+2b+3x, and their difference is a+ 2b-3x; hence:

(a+ 2b)^2-9x^2 = (a+2b+3x)(a+ 2b-3x)

Example 2 Factor m^2 - (2a-3b)^2.

Solution: The sum of m and 2a-3b is m+2a-3b, and their difference is m-(2a-3b) =m-2a+3b. Hence:

m^2-(2a-3b)^2 = (m+2a-3b)(m-2a+3b).

If the factors contain similar terms, they should be collected so as to give the result in its simplest form.

Example 3 Factor (3x + 5y)^2-(2x-y)^2

Solution:

(3x + 5y)^2-(2x-y)^2=[(3x+5y)+(2x-y)][(3x+5y)-(2x-y)]

= (3x+5y+2x-y)(3x+5y-2x+y)

= (5x+4y)(x+ 6y).

### Summary of Factoring

In many problems of factoring it is possible to resolve the given expression into more than two factors. Thus, to factor 2a^2-2b^2 we first remove the common
factor 2, and then factor the remainder since it is the difference of two squares.

2a^2-2b^2 = 2(a^2-b^2) = 2(a + b)(a-b).

In general, to factor more complex expressions we proceed as follows:

1. In all problems of factoring, remove monomial factors before attempting any other method of factoring.

2. Depending on the number of terms in the multinomial that remains, attempt to factor as follows:

(a) If it has two terms, try to factor as the difference of two squares.

(b) If it has three terms, try to factor as a trinomial.

(c) If it has four or more terms, try to factor by grouping.

This process is to be applied to any new factors obtained until every remaining factor can be factored no further.

Example 1. Factor x^3-2x^2-3x.

Solution: This expression contains the common factor x, hence:

x^3-2x^2-3x = x(x^2-2x-3).

Since three terms remain in the second factor, we attempt to factor as a trinomial; and since

x^2-2x-3 = (x+ 1)(x-3)
we have finally:

x^2-2x-3 = x(x+ 1)(x-3).

Example 2 Factor a^4-x^4.

Solution:    a^4-x^4 = (a^2 + x^2)(a^2-x^2)

= (a^2 + x^2)(a + x)(a -x)

Example 3 Factor x^4-7x^2y^2 + 12y^4.

Solution:

x^4-7x^2y^2 + 12y^4 = (x^2-3y^2)(x^2-4y^2)

= (x^2-3y^2)(x + 2y)(x-2y)

Example 4 Factor a^2x^2-a^2y^2-b^x^2+b^2y^2

Solution:

a^2x^2-a^2y^2-b^x^2+b^2y^2=(a^2x^2-a^2y^2)-(b^2x^2-b^2y^2

= a^2(x^2-y^2) - b^2(x^2-y^2)

= (x^2-y^2)(a^2-b^2)

= (x + y)(x-y)(a + b)(a -b)

Example 5 Factor a^2-b^2+a+b.

Solution:     a^2-b^2+a+b= (a^2-b^2) + (a+ b)

= (a + b)(a-b) + (a + b)

= (a+ b)(a-b+1)

### The Sum and Difference of Two Cubes

By direct multiplication one can easily show:

A^3 + B^3 = (A + B)(A^2-AB + B^2)

and     A^3-B^3 = (A-B)(A^2+ AB + B^2).

Using these two relations as formulas, we may factor any binomial which can be written as the sum or the difference of two cubes.

Example 1 Factor 8x^3+27y^3.

Solution: The given expression may be represented as the sum of two cubes in the following manner:

8x^3+ 27y^3 = (2x)^3 + (3y)^3.

Comparing the new expression with the first formula given above we see that A = 2x and B = 3y. Hence substituting these values into the formula:

(2x)^3 + (3y)^3 = [(2x) + (3y)][(2x)^2 - (2x)(3y) + (3y)^2]

= (2x+ 3y)(4x^2- 6y + 9y^2)

Example 2 Factor (2a-1)^3- a^3.

Solution: Comparing the given expression with the second formula given above, we see that A = 2a- 1 and B = a. Hence, substituting these values into the formula:

(2a-1)^3- a^3=[(2a-1)-a][(2a-1)^2+(2a-1)a+a^2

= (a -1)(7a^2 -5a+1).

### The Factor Theorem

One of the most useful methods of factoring depends on the following theorem.

The Factor Theorem
If any polynomial in x becomes equal to 0 when a is written

for x, then the polynomial is exactly divisible by x — a.

Note. A polynomial in x is an expression of the form ax^n +bx^(n-1) + ... px + q, in which the coefficients are constants and n is a positive integer.

Proof: Let P stand for the polynomial. Divide P by x- a until the remainder no longer contains x. Let R denote this remainder, and Q the quotient obtained. Then

P/(x-a)=Q+R/(x-a)

or,           P= Q(x-a)+R

Since this equation is true for all values of z, we will assume that x equals a. By hypothesis, the substitution of a for x makes P equal to 0: thus

0=Q*0+R.

Hence           R= 0.

Since the remainder is zero, the given polynomial is exactly divisible by x-a.

The following examples illustrate the application of this theorem.

Example 1 Factor x^3-x-6.

Solution: By trial we find that this expression reduces to 0 when 2 is substituted for x. Hence, by the Factor Theorem x-2 is a factor. Dividing the given expression by x-2, we obtain the quotient x^2+2x+3. Thus, in factored form we have

x^3-x-6=(x-2)(x^2+2x+3).

Note. In seeking values which will make the given polynomial equal to zero, observe that we test only those numerical values which divide evenly into the constant term of the polynomial. This is true only if the highest power of x in the polynomial has a coefficient equal to +-1 and the other coefficients are integers. We shall restrict ourselves to such polynomials. Thus, in Example 1, we would have tried the values +-1, +-2, +-3, and +-6. If none of these make the polynomial equal to zero, then we may conclude that the polynomial can not be factored by means of the Factor Theorem,

Example 2 Factor x^4+5x- 6.

Solution: The only numbers which divide evenly into the constant term are +-1, +-2, +-3, +-6, and since (+1)^4+5(+1) -6 equals zero we know that x-1 is a factor. Hence, dividing x^4+5x- 6 by x-1,

x^4+5x- 6=(x-1)(x^3+x^2+x+6).

For the new factor, x^3+x^2+x+6, we again seek a zero; and we find that x =-2 makes the expression zero. Thus x - (-2) or x + 2 is a factor, and again by division we find the other factor to be x^2-x+3. Thus far then,

x^4+5x- 6= (x-1)(x+2)(x^2-x+3).

Having reduced the remaining factor to a trinomial, we attempt to factor it as a trinomial, and since it cannot be factored, we conclude that the original expression can be factored no further.