### Natural Number Exponents

In earlier tutorial we defined a polynomial and agreed that when a polynomial expression is written in standard form, each term except the constant term is in the form ax^n, where a is *any* real number but n is a *natural* number. In this tutorial we will investigate more closely expressions, both numerical and algebraic, which have natural number exponents.

Recall that the numerical expression 2^4 means 2∙2∙2∙2. The exponent, 4, indicates how many times the base, 2, is used as a factor. Now consider the product 2^4∙2^3. This is (2∙2∙2∙2)(2∙2∙2), the product of 7 factors, each of which is 2. Then 2^4∙2^3=2^7. Similarly, 3^2∙3^5=(3∙3)(3∙3∙3∙3∙3)=3^8. And 5^5∙5^4=5^9. Note that in each example the exponents are added, while the base remains the same.

**Example 1** If possible, write each expression using only one base and one exponent.

(a) 7^3∙7^5 (b) (-3)^5(-3)^2 (c) (2/3)^2(2/3)^4 (d) 2^3∙3^4

**Solution**

(a) 7^(3+5)=7^8 (b) (-3)^(5+2)=(-3)^7 (c) (2/3)^(2+4)=(2/3)^6 (d) Impossible, since the bases are not the same. But 2^3∙3^4=8∙81=648.

Now consider the algebraic expression x^3∙x^4. Since x^3∙x^4=(x∙x∙x)(x∙x∙x∙x), we see that x^3∙x^4 is equivalent to x^7. Similarly, x^2∙x^10=x^12. More generally, we see that the *product* of any two expressions which have the *same base* and natural number exponents is equivalent to an expression whose base is unchanged and whose exponent is found by *adding* the original exponents. We may express this fact symbolically in the next theorem.

Theorem 1 ;

Ifa EN, and b EN, then 2¢- x = z+

We may use this theorem and our previous axioms about the set of real numbers to simplify many algebraic expressions as the following examples illustrate.

**Example 2 **Find the standard form.

(a) x^5∙x (b) (3x^2)(x^5) (c) (2x^3)(-3x^2)

**Solution**

(a) x^5∙x=x^(5+1)=x^6 (b) (3x^2)(x^5)=3(x^2∙x^5)=3x^7

(c) (2x^3)(-3x^2)=2(-3)(x^3∙x^2)=-6x^5

Often the distributive axiom may be used with Theorem 1.

**Example 3** Find the standard form.

(a) 3x^2(2x+5) (b) (-3x)(x^3-4x^2) (c) 3x^2+2x

**Solution**

(a) 3x^2(2x+5)=(3x^2)(2x)+(3x^2)(5)=6x^3+15x^2

(b) (-3x)(x^3-4x^2)=(-3x)(x^3)+(-3x)(-4x^2)=-3x^4+12x^3

(c) 3x^2+2x is already in standard form. Theorem 1 cannot be used to simplify a sum.

Now consider the numerical expression 2^3∙5^3. We see that 2^3∙5^3=(2∙2∙2)(5∙5∙5)=(2∙5)(2∙5)(2∙5)=(2∙5)^3=10^3. Similarly, 3^4∙7^4=(3∙7)^4=21^4. Notice that when both factors in a *product* have the same exponent, we may *multiply* the bases, but the exponent remains unchanged.

**Example 4** If possible, write each expression using only one base and one exponent.

(a) 5^7∙3^7 (b) (2/3)^4∙3^4 (c) (-3)^5(-2)^5 (d) 5^2∙5^5 (e) 5^3+2^3

**Solution**

(a) (5∙3)^7=15^7 (b) (2/3∙3)^4=2^4

(c) [(-3)(-2)]^5=6^5 (d) 5^(2+5)=5^7

(e) 5^3+2^3!=7^3, But 5^3+ 2^3=125+8=133

The algebraic expression 2^3∙x^3 is equivalent to (2∙2∙2)(x∙x∙x)=(2x)(2x)(2x)=(2x)^3. Similarly, 5^4∙x^4=(5x)^4 and (3x)^7=3^7∙x^7. The *product* of any two expressions which have the same natural number exponent is equivalent to an expression whose exponent remains unchanged and whose base is found by *multiplying* the original bases. This fact is represented symbolically in the next theorem.

Theorem 2 If a EN, then 2x*- y* = (x- y)2

The following examples show how to use Theorem 2 to find equivalent algebraic expressions.

**Example 5** Find an equivalent expression which uses only one exponent.

(a) 4^5x^5 (b) (-3)^2(2x)^2

**Solution**

(a) (4x)^5 (b) (-6x)^2

Often we will want to use Theorem 2 to find the standard form of a polynomial.

**Example 6** Find the standard form.

(a) (2x)^3 (b) (-2x)^2 (c) (-x)^2 (d) (-x)^3

**Solution**

(a) (2x)^3=2^3x^3= 8x^3

(b) (-2x)^2=(-2)^2x^2=4x^2

(c) (-x)^2 = (-1)^2x^2 =1∙x^2=x^2

(d) (-x)^3=(-1)^3x^3=(-1)x^3=-x^3

We may use both Theorems 1 and 2 to simplify polynomials.

**Example 7** Find the standard form.

(a) (3x)^2(2x^3) (b) (-2x)^3(5x^2- x + 3)

**Solution**

(a) (3x)^2(2x^3)=(9x^2)(2x^3)=18x^5

(b) (-2x)^3(5x^2- x + 3)=(-8x^3)(5x^2- x + 3)=-40x^5+8x^4-24x^3

Now consider the numerical expression (2^3)^2. We see that (2^3)^2=(2^3)(2^3)=2^6. Similarly, (5^2)^4 = (5^2)(5^2)(5^2)(5^2) = 5^8. Note that in each case the expression was a power of an expression which was itself a power, and that we *multiplied* the exponents, but did not change the base of the original power.

**Example 8** If possible, write each expression using a single base and a single exponent.

(a) (5^3)^4 (b) [(2/3)^4]^5 (c) [(-2)^3]^5

Solution

(a) 5^(3∙4)=5^12 (b) (2/3)^(4∙5)=(2/3)^20 (c) (-2)^(3∙5)=(-2)^15

Again we may generalize this property of exponents. (x^3)^7 = x^21, [(2x)^2]^5 = (2x)^10, Whenever an expression is a natural number power of an expression which is itself a natural number power, it is equivalent to an expression whose *exponent* is the *product* of the two original exponents, and whose base is the base of the original expression.

Theorem 3 Ifa CN and db EN, then (2)? = x

We shall not try to prove any of them, but we may use the three theorems of this section to find many equivalent expressions.

**Example 9** Find the standard form.

(a) (x^3)^5 (b) [(2x)^3]^2 (c) (3x^2)^3 (d) (-x^3)^2 (e) (-2x^2)^3

**Solution**

(a) x^(3∙5)=x^15 (b) (2x)^(3∙2)=(2x)^6=2^6x^6=64x^6

(c) (3x^2)^3=3^3(x^2)^3=27x^6 (d) (-x^3)^2=(-1)^2(x^3)^2=x^6

(e) (-2x^2)^3=(-2)^3(x^2)^3=-8x^6

### Multiplication of Polynomials

Often we shall want to find the standard form for expressions which contain the product of two polynomials. The distributive axiom, which we can express symbolically by (a + b)c = ac + bc or a(b +c) = ab + ac, and the other assumptions we have made about the set of real numbers can be used to accomplish this as the following example illustrates.

**Example 1** Find the standard form.

(a) (3x-4)(2x) (b) 2x(x+5) + 3(x+5) (c) x(2x-1)-2(2x -1)

**Solution**

(a) (3x-4)(2x)=(3x)(2x)+(-4)(2x)=6x^2-8x

(b) 2x(x+5) + 3(x+5)=2x^2+10x+3x+15=2x^2+13x+15

(c) x(2x-1)-2(2x -1)=x(2x-1)+(-2)(2x -1)=2x^2-x-4x+2=2x^2-5x+2

When each factor of a product has two or more terms, we may still use the distributive axiom to find the standard form, but the problem is considerably more complicated. For example, we can treat the product (2x + 3)(3x + 4) as if (2x + 3) is considered as a sum (a + b) and (3x + 4) is considered as a single factor, c. Then (a + b)c = ac + bc becomes (2x + 3)(3x + 4) = (2x) (3x + 4)+(3)(3x + 4). Each of these products can be simplified again using the distributive axiom. 2x(3x + 4)= 6x^2 + 8x and 3(3x + 4) = 9x + 12. The standard form is then obtained by adding similar terms. 6x^2 + 8x + 9x + 12 = 6x^2 + 17x + 12.

That is, (2x + 3)(3x + 4) = 2x(3x + 4) + 3(3x + 4) = 6x^2 + 8x + 9x + 12 = 6x^2 + 17x + 12(2x + 3)(3x + 4) = 2x(3x + 4) + 3(3x + 4) = 6x^2 + 8x + 9x + 12 = 6x^2 + 17x + 12. In fact every product of polynomials may be written in standard form by repeated use of the distributive axiom. Some examples will further illustrate the technique of doing so.

**Example 2** Find the standard form.

(a) (2x^2 + 3)(x-2) (b) (4x -5)(3x + 2) (c) (2x-3)(x^2 + 4x-1) (d) (x^2 + 2x -1)(2x + 5)

**Solution**

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

(b) (4x -5)(3x + 2)=4x(3x+2)+(-5)(3x+2)=12x^2+8x-15x-10=12x^2-7x-10

(c) (2x-3)(x^2 + 4x-1)=2x(x^2 + 4x-1)+(-3)(x^2 + 4x-1)=2x^3+8x^2-2x-3x^2-12x+3

=2x^3+5x^2-14x+3

(d) (x^2 + 2x -1)(2x + 5)=(2x + 5)(x^2 + 2x -1)=2x(x^2 + 2x -1)+5(x^2 + 2x -1)

=2x^3+4x^2-2x+5x^2+10x-5

=2x^3+9x^2+8x-5

To allow us to refer to polynomials in a convenient way we shall occasionally use the expressions trinomial, binomial, or monomial. A polynomial such as 3x^2- 7x + 5 or 4x^5-2x^2 + 3x which has three terms we shall call a **trinomial**. A polynomial such as 2x + 3, x^2-7 and 3x^2 + 4x which has exactly two terms we shall call a **binomial**. Polynomials with fewer than two terms such as 4x^3, -2x, and 17 we shall sometimes call **monomials**. Polynomials with more than three terms will not be given any special names. Then (2x + 5)(3x + 4) is the product of two first degree binomials. We shall want to find the standard form for such a product so often that it will be useful for us to develop a schematic method which will somewhat shorten the process by allowing us to do much of the work mentally.

Let us make several observations about the standard form of the polynomial in Figure 4.1. Notice that the product of the two *first* degree binomials 2x + 5 and 3x + 4 is equivalent to the *second* degree trinomial 6x^2 + 23x + 20. The second degree *term* 6x^2 is merely the product of 2x and 3x, the first degree *terms* of the factors. And the constant term 20 is the product of the constant terms of the factors. But the first degree term 23x is the *sum* of two other first degree terms, one of which is the product of 5 and 3x, the other of which is the product of 2x and 4. These observations are quite generally true for the product of two first degree binomials. Let us summarize them now and illustrate them with some examples.

The product of two first degree binomials is always a second degree

polynomial whose:

1. Second degree term is the product of the two first degree terms of

the factors.

2. First degree term is found by adding two first degree terms each of

which is the product of the constant term from one factor and the

first degree term from the other factor.

3. Constant term is the product of the two constant terms of the

factors.

**Example 3** Find the standard form and show schematically how the first degree term is found.

(a) (x + 3)(2x + 5) (b) (2x -1)(5x + 2) (c) (4x - 3)(2x -5)

**Solution**

5x

TF

(a) (x + 3)(22 + 5) = 227? 4+ Ilex + 15 (2 + 3)(2x + 5)

6x

4x

7

(b) (2x — 1)(5a + 2) = 10x? —2 — 2 (2x — 1)(5x + 2)

=

—5x

— 20x

7

(c) (4% — 3)(2z — 5) = 8x? — 267 4+ 15 (42 — 8) (2x — 5)

— 6x

It will usually not be necessary to show schematically how the product is obtained. When all computations are done mentally, we shall say the result was obtained by inspection.

**Example 4** Find the standard form by inspection.

(a) (4x-5)(3x + 1) (b) (2x-5)(x + 4) (c) (x + 4)(x-4)

**Solution**

(a) 12x^2-11x-5 (b) 2x^2 + 3x-20 (c) x^2-16

In the last example we have (x + 4)(x-4)=x^2-16. Notice that the first degree term of the standard form is the sum of 4x and -4x, which is 0. It is generally true that one of the terms will be zero when we multiply two binomials with one factor the sum of two terms and the other factor the difference of *those same two terms*. The next theorem expresses this fact symbolically.

Theorem 4 (a + b)(a —b) =a? —B

Proof (a + b)(a — b) = a(a — b) + b(a — d)

= a’? — ab + ba — b? = a? — 9?

We may express Theorem 4 verbally by saying that the product of the sum of two terms and the *difference* of those same two terms is equivalent to the difference of the square of the first term and the square of the second term. And we need not restrict ourselves to products of first degree binomials, although we often shall.

**Example 5** Use Theorem 4, if possible, to find the standard form.

(a) (x + 5)(x -5) (b) (2x + 3)(2x -3) (c) (4x^2-5)(4x^2+5) (d) (x+4)(x-3)

**Solution**

(a) x^2-25 (b) 4x^2-9 (c) 16x^4-25 (d) Impossible, but (x + 4)(x-3) = x^2+x-12

Often, we will want to find the standard form of the square of a binomial.

**Example 6** Find the standard form of (2x + 3)^2.

**Solution** (2x + 3)^2 = (2x + 3)(2x + 3) = 4x^2 + 12x + 9. Notice that the standard form of the square of the binomial 2x + 3 is a trinomial whose second degree term 4x^2 is the square of the first degree term 2x, and whose constant term 9 is the square of the constant term 3. But the first degree term of the trinomial, 12x, is twice the product of the first degree term 2x and the constant term 3. That is, 2[(2x)(3)] = 12x. The next theorem shows that this is generally a true result.

Theorem 5 (a +b)? = a? + 2ab +

Proof (a+ b)? = (a+ b)(a+b) = a(a+b) + b(a + b)

= a’+ ab + ba + b? = a? + 2ab + b?

**Example 7** Use Theorem 5, if possible, to find the standard form by inspection.

(a) (x + 2)^2 (b) (3x + 5)^2 (c) (2x-3)^2

**Solution**

(a) x^2+4x+4 (b) 9x^2 + 30x + 25 (c) 4x^2-12x + 9. Note that twice the product of the two terms of the binomial is 2[(2x)(-3)] = -12x.

The theorems and techniques of this section may be used to find the standard form for a variety of polynomials.

**Example 8** Find the standard form.

(a) (2x)(3x-1)(x + 5) (b) (3x + 2)(x + 5)(3x -2)

(c) (x-4)(2x-1)^2 (d) [2x(x -3)]^2

**Solution**

(a) (2x)(3x-1)(x + 5)=2x[(3x-1)(x+5)]=2x(3x^2+14x-5)=6x^3+28x^2-10x

(b) (3x + 2)(x + 5)(3x -2)=(x+5)[(3x+2)(3x-2)]=(x+5)(9x^2-4)=x(9x^2-4)+5(9x^2-4)

=9x^3-4x+45x^2-20=9x^3+45x^2-4x-20

(c) (x-4)(2x-1)^2=(x-4)(4x^2-4x+1)=x(4x^2-4x+1)+(-4)(4x^2-4x+1)

=4x^3-4x^2+x-16x^2+16x-4= 4x^3-20x^2+17x-4

(d) [2x(x -3)]^2=(2x)^2(x-3)^2=4x^2(x^2-6x+9)=4x^4-24x^3+36x^2

### Factoring Polynomials

When an expression is written as a product, we shall say that it is in factored form. Then (x + 2)(x -3), -14x^2, 3x(x + 2) and (2x-1)^2 are all in factored form, but neither 3x + 4 nor 2x(x + 5) + 3 is in factored form.

**Example 1** Which of these expressions are in factored form?

(a) (4x + 5)(4x-5) (b) 2x(3x -1)^2 (c) 2x^2 (d) 3x(x + 2) + 5(x + 2) (e) x(x +5) -3

**Solution** Only (a), (b), and (c) are in factored form.

Often we will want to replace an expression which is *not* in factored form by an equivalent expression which *is* in factored form, and we shall call this process "factoring the expression." The distributive axiom may often be used to factor an expression.

**Example 2** Factor these expressions.

(a) 3x + 6 (b) 2x -8 (c) x^2 + 3x

**Solution**

(a) 3(x + 2) (b) 2(x-4)]] (c) x(x + 3)

Notice that each of the expressions in Example 2 was factored by finding a factor which was repeated in each term. The repeated factor in 3x + 6 is 3, since 3 is a factor of both 3x and 6; the repeated factor in 2x-8 is 2; and in x^2 + 3x it is x. Finding a factor which is repeated in each term is a commonly used technique for factoring an expression.

**Example 3 ** Factor each expression by finding a factor which is repeated in each term.

(a) 5x^2 + 10x-20 (b) 2x^3 -5x^2+ x (c) x(x + 1) + 2(x +1)

**Solution**

(a) 5(x^2 + 2x-4) (b) x(2x^2 -5x+ 1 (c) The repeated factor is the binomial x + 1. Then x(x + 1) + 2(x +1)=(x+2)(x+1)

Sometimes Theorem 4, (a+b)(a-b) = a^2-b^2, may be used to factor an expression even when there is no repeated factor. If a *binomial* is the difference of squares, we may always factor it.

**Example 4** Factor, if possible, using Theorem 4.

(a) x^2-9 (b) 4x^2-1 (c) 16x^2- 25 (d) x^4-4 (e) x^2 + 49

**Solution**

(a) (x + 3)(x-3) (b) (2x + 1)(2x-1)

(c) (4x + 5)(4x-5) (d) (x^2 + 2)(x^2-2)

(e) Impossible, since x^2 + 49 is not the *difference* of squares. Indeed, the only ways we shall be able to factor a binomial are the removal of a repeated factor and by means of Theorem 4.

However, Theorem 5, (a + b)^2 = a^2 + 2ab + b^2, can be used to factor a trinomial which is the square of a binomial. For example, the trinomial x^2 + 6x + 9 is the square of x + 3. Notice that its first term x^2 is the square of x and its constant term 9 is the square of 3. This suggests that it *might* be the square of x + 3. And in fact (x + 3)^2 = x^2+ 6x+9. We can also factor the trinomial x^2-10x + 25 by noting that the first and third terms, x^2 and 25, are the squares of x and 5, respectively. And while (x + 5)^2 = x^2 + 10x + 25 is not the result we seek, (x-5)^2 = x^2-10x + 25, and so the factored form is (x -5)^2.

**Example 5 **Use Theorem 5, if possible, to factor each of the following:

(a) 4x^2+12x+9 (b) 9x^2-30x+25

(c) x^2+49 (d) x^2+6x-9

(e) x^2+4x+9 (f) x^2+4x+3

**Solution**

(a) 4x^2 is the square of 2x and 9 is the square of 3

(2x+3)^2 = 4x^2+12x+9

(b) 9x^2 is the square of 3x and 25 is the square of -5

(3x -5)^2 = 9x^2-30x+25

(c) Theorem 5 can not be used to factor x^2 + 49, since it is not a trinomial.

(x+7)^2!=x^2+49. In fact (x + 7)^2 = x^2 + 14x + 49

(d) Theorem 5 can not be used to factor x^2+6x-9. The constant term -9 is not the square of any real number.

(e) Theorem 5 can not be used to factor x^2+4x+9. While x^2 is the square of x and 9 is the square of 3, (x + 3)^2 = x^2 + 6x + 9, which is not the given trinomial.

(f) Theorem 5 can not be used to factor x^2+4x+3, since 3 is not the square of an integer.

While x^2 + 4x + 3 cannot be factored by any of the techniques so far discussed, it is easily seen that x^2 + 4x + 3 = (x + 1)(x + 3). Let us investigate a method which can be used to find factors such as these when it is possible to do so.

We shall first try to factor x^2 + 8x + 12. This is a second degree trinomial which has no repeated factors and is not the square of a binomial. We shall try to find a pair of first degree binomial factors. Recall that the second degree term x^2 must be the product of the two first degree terms of the factors. Hence, these terms must both be x. And the constant term 12 must be the product of the two constant terms of the factors. However, there are several choices for the constant terms: 3 and 4, 2 and 6, 1 and 12. We shall try 3 and 4 as a first guess. But (x+ 3)(x + 4) = x^2+ 7x+ 12, and so we have *not* found the factored form. Next we shall try 2 and 6. Since (x + 2)(x + 6) = x^2 + 8x + 12, we have found the factored form.

**Example 6** Factor 3x^2 + 7x + 4.

**Solution** The first degree terms must be 3x and x since (3x)(x) = 3x^2]]. The constant terms may be 2 and 2, or 1 and 4. We begin to make trials.

(3x + 2)(x + 2) = 3x^2+ 8x+ 4

(3x + 1)(x+ 4) = 3x^2+ 13x +4

(3x + 4)(x + 1) = 3x^2 + 7x + 4

Hence the factored form is (3x + 4)(x + 1).

When the first degree term of the trinomial has a negative coefficient, further possibilities must be considered. The constant term may be positive as the following example illustrates.

**Example 7** Factor 4x^2- 8x + 3.

**Solution** The first degree terms may be 4x and x or 2x and 2x. Since the first degree term of the trinomial is -8x, the constant terms of the factors must be negative, and we will try -1 and -3.

(4x-1)(x-3) = 4x^2- 13x + 3

(4x-3)(x-1) = 4x^2-7x + 3

(2x- 3)(2x-1) = 4x^2- 8x+ 3

Hence the factored form is (2x-3)(2x-1).

Or the constant term of the trinomial may be negative, requiring that even more possibilities be considered.

**Example 8** Factor x^2 + x -12.

**Solution** The first degree terms must be both x. There are several possibilities for the constant terms: 3 and -4, -3 and 4, 2 and -6, -2 and 6, 1 and -12, -1 and 12. By trial and error we hope to find the required factors.

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

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

Our second trial has proved successful and the factored form is (x-3)(x + 4). But note that we might have tried many other possibilities before finding the correct one.

As the examples illustrate, factoring a second degree trinomial is a process of trial and error, but we may often limit the trials to only a few. It will be helpful to summarize these limiting conditions on the terms of the binomial factors.

When factoring a nonsquare second degree trinomial into the product

of two first degree binomials:

1. The product of the first degree terms of the factors must equal the

second degree term of the trinomial.

2. The product of the constant terms of the factors must equal the

constant term of the trinomial.

3. When the constant term of the trinomial is positive, the constant

terms of the factors will be both positive or both negative, depend-

ing on whether the coefficient of the first degree term of the trinomial

is positive or negative.

4. When the constant term of the trinomial is negative, one constant

term of the factors will be positive and the other will be negative.

**Example 9** Factor each of the following trinomials:

(a) 5x^2-31x+6 (b) 4x^2+15x-4

**Solution**

(a) 5x^2-31x+6. The factors of the first term are 5x and x. The factors of the third term might be -3 and -2 or -6 and -1. (We use negative constants since the coefficient of the second term in the trinomial is negative.) There are four cases to consider:

(5x-3)(x-2) = 5x^2-13x+6

(5x-2)(x-3) = 5x^2-17x+6

(5x-6)(x-1) = 5x^2-11x+6

(5x-1)(x-6) = 5x^2-31x+6

The last pair of factors is the correct one.

(b) 4x^2+15x-4. The factors of the first term might be 2x and 2x, or 4x and x. The factors of the constant term might be 2 and -2, 4 and -1 or -4 and 1. There are actually nine possibilities to consider.

Trial and error eventually yield the correct factors.

4x^2+15x-4=(4x-1)(x+4)

We end this section with a summary of the methods of factoring we have considered.

1. When each term of a binomial or a trinomial contains a repeated

factor, we can remove this repeated factor.

4x? — 62 + 18 = 2(22? — 3x + 9).

2. When a binomial is the difference of two squares, we can use

Theorem 4 to obtain its two factors. 422 — 49 = (22 + 7)(2% — 7).

3.hen a trinomial is the square of a binomial, Theorem 5 gives its

factors. 9x? — 242 + 16 = (3¢ — 4)2.

4. When a trinomial of degree 2 is not a square of a binomial, it may

be possible to find two first degree binomials which are its factors.

6z? — 7x — 3 = (2x — 3)(8z2 + 1).

In the next section we shall look more closely at the subject of factoring. We shall observe that some polynomials can be factored in more than one way, some polynomials cannot be factored at all in ways which are useful to us, and there is a standard form of factored expressions.

### Complete Factoring

Sometimes it is possible to factor an expression in more than one way. For example, the binomial 4x^2+6x may be factored in at least three ways: 2(2x^2 + 3x), x(4x + 6) and 2x(2x + 3). In this section we will look more closely at factoring and agree on a best or most complete factorization.

We have learned two methods for factoring a binomial. First, we may remove a common factor: 2x + 4 = 2(x + 2). Second, we may factor the difference of squares: x^2-1 = (x + 1)(x-1). There are no other techniques for factoring a binomial which we will consider. Then is it possible to factor 2x + 1? Surely this is not the difference of squares, and there would seem to be no repeated factor. But notice that 2x + 1 = 2(x + 1/2). Have we factored the expression 2x + 1? Until now we have considered only polynomials in which all of the coefficients (and the constant term) are *integers*. Notice that 2x + 1 is such a binomial with integral coefficients, but that x + 1/2 is not, since 1/2 is not an integer. Let us agree that the factored form of a polynomial with integral coefficients shall be a product of polynomials all of whose coefficients are integers. Then we will not consider 2(x+1/2) to be the factored form of 2x + 1. But what is the factored form of 2x + 1? It happens that with the restrictions we have imposed about integral coefficients, the only factored forms for 2x + 1 are (1)(2x + 1) and the equally trivial (-1)(-2x-1). We will not usually be interested in either of these trivial factorizations, but will say that 2x + 1 is a **prime** polynomial, and shall not attempt to factor it. More generally we will say that a polynomial is prime whenever its only allowable factorizations are the two trivial ones.

Let us return to the factorization of 4x^2 + 6x. We noted that 2(2x^2 + 3x), x(4x + 6) and 2x(2x + 3) are all factorizations of 4x^2 + 6. Notice that only in the factorization 2x(2x + 3) is the binomial factor 2x + 3 a prime. For both 2x^2 + 3x and 4x + 6 may be factored and so are not prime. Let us agree that 2x(2x + 3) is the complete factorization of 4x^2 + 6x, since the binomial factor 2x + 3 is prime. More generally, an expression will be considered completely factored only if each of its factors (which are not monomials) are prime polynomials.

**Example 1** Find the complete factorization.

(a) 12x^2-16x (b) 5x^3+10x^2

**Solution** (a) 4x(3x-4) (b) 5x^2(x+2)

Sometimes the complete factorization of a binomial] may be found by first removing a common factor and then factoring the difference of squares. The next example illustrates this.

**Example 2** Find the complete factorization

(a) x^3-x (b) 2x^3-50x (c) x^4-16

**Solution**

(a) x^3-x=x(x^2-1)=x(x+1)(x-1). Notice that x^2-1 was not prime, so we replaced it with (x + 1)(x-1).

(b) 2x^3-50x=2x(x^2-25)=2x(x+5)(x-5)

(c) x^4-16=(x^2+4)(x^2-4)=(x^2+4)(x+2)(x-2). Notice that while x^2-4 is the difference of Squares and may be factored, x^2 +4 is not and is a prime polynomial.

We have learned three techniques for factoring a trinomial. First, by removing a common factor:

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

Second, as the square of a binomial:

x^2+4x+4=(x+2)^2

Third, as the product of two different binomials:

x^2+ 3x -4 =(x + 4)(x -1)

These are the only techniques we will consider, and as is the case for binomials, we may have to use two or more of these methods to find the complete factorization of a trinomial.

**Example 3 **Factor completely.

(a) 2x^3-12x^2 + 18x (b) 6x^2+3x-9

**Solution**

(a) 2x^3-12x^2 + 18x=2x(x^2-6x+9=2x(x-3)^2

(b) 6x^2+3x-9=3(2x^2+x-3)=3(2x+3)(x-1)

Sometimes both binomials and trinomials may be more easily factored if we first remove the trivial factor -1. We will want to do this whenever the coefficient of the highest degree term is negative.

**Example 4** Factor completely.

(a) -x^2+16 (b) -x^2+4x-4 (c) -3x^2-3x+18

**Solution**

(a) -x^2+16=(-1)(x^2-16)=-(x+4)(x-4)

(b) -x^2+4x-4=-(x^2-4x+4)=-(x-2)^2

(c) -3x^2-3x+18=-3(x^2+x-6)=-3(x+3)(x-2)

When the polynomial is not in standard form we will often (but not always) want to first write it in standard form before factoring.

**Example 5** Factor completely.

(a) x(x -2) - 3 (b) 3x(x + 1) + 6(x + 1)

**Solution**

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

(b) Since there is the repeated factor x + 1, we will not write the polynomial in standard form.

3x(x + 1) + 6(x + 1)=(3x+6)(x+1)=3(x+2)(x+1)

Notice that 3x + 6 was not prime, so we replaced it with 3(x + 2).

Usually we will be unable to factor an expression with more than three terms, but there are two techniques which may sometimes be used. Of course, we may remove a common monomial factor:

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

And sometimes we may group the terms and thus find a repeated factor. Example 6 illustrates this method.

**Example 6** Factor x^3 + x^2 + 2x + 2.

**Solution** We shall group the first two terms and the last two terms.

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

Sometimes we will want to factor a polynomial with rational coefficients, some of which are not integers, such as x + 1/2. Notice that (1/2)(2x + 1) = x +1/2, and so (1/2)(2x + 1) is a factorization. And notice that the binomial factor 2x + 1 has *integral* coefficients. Let us agree that the complete factorization of a polynomial with rational coefficients will be a product in which every factor except the monomial factor is a prime polynomial with integral coefficients.

**Example 7** Factor completely.

(a) 1/3x + 2 (b) 1/2x^2-2

**Solution**

(a) 1/3x + 2=(1/3)(x+6)

(b) 1/2x^2-2=(1/2)(x^2-4)=(1/2(x+2)(x-2)

Let us now review all of the techniques we have considered for factoring polynomials.

1. For any polynomial we will first remove any factor which is repeated

in each term. When the coeffici the highes

polynomial is negative, we will remove the common factor —1.

2. If a binomial is the difference of squares, we will factor it by Theorem

4:a? — b? = (a + b)(a — BD).

3. If a trinomial is the square of a binomial, we will factor it by

Theorem 5: a? + 2ab + b? = (a + 6)?.

4, If a trinomial is the product of two different binomials, we will

factor it by trial and error.

5. If a polynomial has more than three terms, we will group the terms

and try to find a repeated factor.

- __-- «326. When a polynomial has rational coefficients which are not integers,

4 | we will first factor it into the product of a rational number and a

polynomial with integral coefficients and then proceed to factor this

latter polynomial with the rules stated above.

J

Let us agree that to “factor” shall mean to “factor completely,” and the complete factorization of a polynomial shall be a factorization in which every factor except the monomial factor is a prime polynomial with integral coefficients. Some examples will illustrate these rules for factoring.

**Example 8** Factor each of the following.

(a) -4x^2 + 100 (b) 3x^3-24x^2+48x

(c) x^2-5/2x+3/2 (d) 2x^3-x^2-8x+4

(e) 2x^2+3x+2

**Solution**

(a) -4x^2 + 100=-4(x^2-25 [Rule 1]

= -4(x + 5)(x- 5) [Rule 2]

(b) 3x^3-24x^2+48x=3x(x^2-8x+16 [Rule 1]

= 3x(x-4)^2 [Rule 3]

(c) x^2-5/2x+3/2=(1/2)(2x^2-5x+3) [Rule 6]

= (1/2)(2x- 3)(x- 1) [Rule 4]

(d) 2x^3-x^2-8x+4=(2x^3-x^2)+(-8+4x)

= x^2(2x-1) + (-4)(2x-1)

= (x^2-4)(2x - 1) [Rule 5]

= (x + 2)(x-2)(2x-1) [Rule 2]

(e) 2x^2+3x+2 cannot be factored by any of the given rules. It is a prime polynomial.