5x 2 11x 2 Factor

PRODUCTS AND FACTORS

THE DISTRIBUTIVE Constabulary

If we want to multiply a sum by another number, either we can multiply each term of the sum by the number earlier we add or nosotros can first add the terms and and then multiply. For instance,

In either case the result is the same.

This property, which we kickoff introduced in Section 1.8, is called the distributive police force. In symbols,

a(b + c) = ab + ac or (b + c)a = ba + ca

By applying the distributive law to algebraic expressions containing parentheses, we can obtain equivalent expressions without parentheses.

Our first example involves the product of a monomial and binomial.

Case 1 Write 2x(x - iii) without parentheses.

Solution

We think of 2x(10 - 3) as 2x[10 + (-3)] and then utilize the distributive police to obtain

The above method works every bit besides with the product of a monomial and trinomial.

Example 2 Write - y(yⁱⁱ + 3y - 4) without parentheses.

Solution

Applying the distributive property yields

When simplifying expressions involving parentheses, we first remove the parentheses and then combine like terms.

Example iii Simplify a(3 - a) - two(a + a²).

We brainstorm by removing parentheses to obtain

Now, combining like terms yields a - 3a².

We tin use the distributive belongings to rewrite expressions in which the coefficient of an expression in parentheses is +1 or - 1.

Case 4 Write each expression without parentheses.
a. +(3a - 2b)
b. -(2a - 3b)

Solution

Observe that in Example 4b, the sign of each term is changed when the expression is written without parentheses. This is the same event that nosotros would have obtained if we used the procedures that we introduced in Section 2.5 to simplify expressions.

FACTORING MONOMIALS FROM POLYNOMIALS

From the symmetric belongings of equality, we know that if

a(b + c) = ab + ac, then ab + ac = a(b + c)

Thus, if there is a monomial gene common to all terms in a polynomial, we can write the polynomial as the product of the common gene and some other polynomial. For instance, since each term in x² + 3x contains x as a factor, nosotros can write the expression as the product x(ten + 3). Rewriting a polynomial in this mode is called factoring, and the number ten is said to be factored "from" or "out of' the polynomial x² + 3x.

To factor a monomial from a polynomial:

Write a set of parentheses preceded by the monomial common to each term in the polynomial.
Divide the monomial factor into each term in the polynomial and write the quotient in the parentheses.

More often than not, we can observe the common monomial factor past inspection.

Example 1
a. 4x + 4y = 4(x + y)
b. 3xy -6y - 3y(x - 2)

We can check that we factored correctly by multiplying the factors and verifying that the production is the original polynomial. Using Example 1, nosotros go

If the common monomial is hard to find, nosotros tin write each term in prime factored form and note the common factors.

Example two Factor 4x³ - 6x² + 2x.

Solution
We can write

We now encounter that 2x is a common monomial factor to all three terms. Then we gene 2x out of the polynomial, and write
2x( )

Now, we divide each term in the polynomial by 2x

and write the quotients within the parentheses to become

2x(2x² - 3x + 1)

We tin check our answer in Instance 2 by multiplying the factors to obtain

In this volume, we will restrict the common factors to monomials consisting of numerical coefficients that are integers and to integral powers of the variables. The choice of sign for the monomial factor is a matter of convenience. Thus,

-3x² - 6x

can be factored either as

-3x(x + two) or equally 3x(-ten - ii)

The first form is commonly more convenient.

Case 3 Factor out the common monomial, including -i.

a. - 3xⁱⁱ - 3 xy
b. -x³ - xⁱⁱ + x
Solution

Sometimes it is convenient to write formulas in factored class.

Example four
a. A = P + PRT
= P(1 + RT)

b. S = 4kR² - 4krⁱⁱ
= 4k(R² - r²)

iv.3 BINOMIAL PRODUCTS I

We can employ the distributive police force to multiply two binomials. Although there is little need to multiply binomials in arithmetic every bit shown in the example beneath, the distributive law likewise applies to expressions containing variables.

We volition at present apply the above process for an expression containing variables.

Instance 1

Write (x - 2)(ten + 3) without parentheses.

Solution
Get-go, utilize the distributive holding to go

Now, combine similar terms to obtain
x² + x - half-dozen

With practice, yous will be able to mentally add the second and third products. Theabove process is sometimes called the FOIL method. F, O, I, and L stand for:

1. The production of the First terms.
two. The production of the Outer terms.
three. The production of the Inner terms.
four. The product of the Terminal terms.

The FOIL method tin too be used to foursquare binomials.

Case ii

Write (x + three)^two without parentheses.
Solution

First, rewrite (x + 3)^two equally (ten + 3)(10 + 3). Side by side, apply the FOIL method to get

Combining similar terms yields
ten² + 6x + 9

When we have a monomial factor and two binomial factors, it is easiest to first multiply the binomials.

Example iii

Write 3x(10 - 2)(x + three) without parentheses.
Solution
First, multiply the binomials to obtain
3x(x² + 3x - 2x - 6) = 3x(x^two + ten - 6)

At present, apply the distributive police force to get
3x(ten² + x - half dozen) = 3x³ + 3x² - 18x

Mutual Errors

Detect in Case 2

Similarly,

In general,

four.four FACTORING TRINOMIALS I

In Section 4.3, we saw how to find the product of ii binomials. At present we volition reverse this process. That is, given the production of ii binomials, we will notice the binomial factors. The process involved is another case of factoring. As before,we will only consider factors in which the terms have integral numerical coefficients. Such factors do not ever exist, simply we will study the cases where they do.

Consider the following production.

Notice that the first term in the trinomial, ten², is product (i); the last term in the trinomial, 12, is product and the centre term in the trinomial, 7x, is the sum of products (two) and (iii). In full general,

Nosotros use this equation (from right to left) to factor any trinomial of the form x² + Bx + C. Nosotros notice two numbers whose production is C and whose sum is B.

Instance 1 Factor x^two + 7x + 12.

Solution
Nosotros look for two integers whose production is 12 and whose sum is 7. Consider the following pairs of factors whose production is 12.

We see that the only pair of factors whose product is 12 and whose sum is 7 is 3 and 4. Thus,

x² + 7x + 12 = (x + 3)(ten + 4)

Note that when all terms of a trinomial are positive, nosotros need only consider pairs of positive factors because nosotros are looking for a pair of factors whose production and sum are positive. That is, the factored term of

x^two + 7x + 12 would exist of the form

( + )( + )

When the get-go and third terms of a trinomial are positive but the heart term is negative, we need simply consider pairs of negative factors because nosotros are looking for a pair of factors whose product is positive but whose sum is negative. That is,the factored grade of

x² - 5x + six

would be of the course

(-)(-)

Case two Factor x^two - 5x + vi.

Solution
Considering the third term is positive and the middle term is negative, we find two negative integers whose product is half-dozen and whose sum is -5. We list the possibilities.

We see that the only pair of factors whose product is 6 and whose sum is -v is -3 and -2. Thus,

x² - 5x + 6 = (ten - iii)(10 - 2)

When the first term of a trinomial is positive and the third term is negative,the signs in the factored form are opposite. That is, the factored grade of

10ⁱⁱ - x - 12

would exist of the form

( + )( - ) or ( - )( + )

Case 3

Factor 10² - x - 12.

Solution We must find ii integers whose product is -12 and whose sum is -1. We list the possibilities.

Nosotros encounter that the only pair of factors whose production is -12 and whose sum is -1 is -four and iii. Thus,

x² - x - 12 = (x - 4)(ten + 3)

It is easier to cistron a trinomial completely if any monimial factor common to each term of the trinomial is factored first. For example, we can factor

12x² + 36x + 24

A monomial can so be factored from these binomial factors. However, first factoring the common gene 12 from the original expression yields

12(x² + 3x + 2)

Factoring again, nosotros take

12(* + 2)(x + i)

which is said to be in completely factored course. In such cases, it is not necessary to gene the numerical factor itself, that is, nosotros do not write 12 every bit 2 * two * iii.

Example 4

Gene 3x² + 12x + 12 completely.

Solution
First we factor out the 3 from the trinomial to become

3(ten² + 4x + 4)

At present, we gene the trinomial and obtain

3(x + ii)(10 + 2)

The techniques we have developed are also valid for a trinomial such every bit ten² + 5xy + 6y².

Example fiveFactor x² + 5xy + 6y².

Solution
We find two positive factors whose product is 6y² and whose sum is 5y (the coefficient of 10). The 2 factors are 3y and 2y. Thus,

x² + 5xy + 6y² = (10 + 3y)(x + 2y)

When factoring, it is all-time to write the trinomial in descending powers of x. If the coefficient of the x²-term is negative, factor out a negative before proceeding.

Example 6

Factor 8 + 2x - 10².

Solution
We first rewrite the trinomial in descending powers of x to get

-x² + 2x + 8

Now, nosotros can cistron out the -ane to obtain

-(x^two - 2x - 8)

Finally, we gene the trinomial to yield

-(x- 4)(x + 2)

Sometimes, trinomials are not factorable.

Example vii

Factor x² + 5x + 12.

Solution
We look for two integers whose product is 12 and whose sum is 5. From the table in Case 1 on page 149, we see that there is no pair of factors whose product is 12 and whose sum is v. In this instance, the trinomial is not factorable.

Skill at factoring is usually the result of all-encompassing practice. If possible, do the factoring process mentally, writing your answer directly. You lot can cheque the results of a factorization by multiplying the binomial factors and verifying that the product is equal to the given trinomial.

iv.five BINOMIAL PRODUCTS II

In this section, we apply the process developed in Section 4.3 to multiply binomial factors whose offset-caste terms take numerical coefficients other than ane or - 1.

Example 1

Write as a polynomial.

a. (2x - 3)(10 + 1)
b. (3x - 2y)(3x + y)

Solution

We first apply the FOIL method so combine similar terms.

As before, if we have a squared binomial, we kickoff rewrite it as a production, then apply the FOIL method.

Case 2

a. (3x + two)²
= (3x + 2)(3x + 2)
= 9x^two + 6x + 6x + iv
= 9x^two + 12x + 4

b. (2x - y)^two
= (2x - y)(2x - y)
= 4x² - 2xy - 2xy + yⁱⁱ
- 4x^two - 4xy + y^two

As yous may have seen in Section four.3, the production of 2 bionimals may have no first-degree term in the reply.

Example 3

a. (2x - 3)(2x + 3)
= 4x² + 6x - 6x - 9
= 4x² -9

b. (3x - y)(3x + y)
- 9x² + 3xy - 3xy - y² = 9xⁱⁱ - y²

When a monomial factor and two binomial factors are being multiplied, it is easiest to multiply the binomials first.

Instance 4

Write 3x(2x - l)(ten + two) as a polynomial.

Solution
Nosotros first multiply the binomials to get 3x(2x² + 4x - 10 - 2) = 3x(2x^two + 3x - two)
Now multiplying by the monomial yields 3x(2x²) + 3x(3x) + 3x(-2) = 6x^three + 9x^two - 6x

four.half dozen FACTORING TRINOMIALS Ii

In Section 4.4 we factored trinomials of the form 10² + Bx + C where the second- degree term had a coefficient of one. Now nosotros desire to extend our factoring techniques to trinomials of the form Axⁱⁱ + Bx + C, where the second-degree term has a coefficient other than ane or -1.

First, we consider a test to determine if a trinomial is factorable. A trinomial of the form Ax² + Bx + C is factorable if we can find two integers whose product is A * C and whose sum is B.

Example 1

Determine if 4x^two + 8x + iii is factorable.

Solution
We check to see if at that place are 2 integers whose production is (4)(3) = 12 and whose sum is 8 (the coefficient of x). Consider the following possibilities.

Since the factors 6 and 2 have a sum of 8, the value of B in the trinomial Ax² + Bx + C, the trinomial is factorable.

Case 2

The trinomial 4xⁱⁱ - 5x + 3 is not factorable, since the in a higher place table shows that there is no pair of factors whose product is 12 and whose sum is -five. The test to see if the trinomial is factorable can usually exist done mentally.

Once we have determined that a trinomial of the course Ax² + Bx + C is fac- torable, we proceed to observe a pair of factors whose product is A, a pair of factors whose product is C, and an arrangement that yields the proper middle term. Nosotros illustrate by examples.

Example three

Factor 4x² + 8x + iii.

Solution
Higher up, we adamant that this polynomial is factorable. Nosotros now proceed.

one. We consider all pairs of factors whose product is 4. Since 4 is positive, merely positive integers need to exist considered. The possibilities are 4, ane and ii, ii.
2. We consider all pairs of factors whose product is 3. Since the middle term is positive, consider positive pairs of factors only. The possibilities are 3, 1. We write all possible arrangements of the factors as shown.

iii. We select the system in which the sum of products (2) and (three) yields a middle term of 8x.

Now, we consider the factorization of a trinomial in which the constant term is negative.

Case 4

Factor 6x^two + x - two.

Solution
First, we test to see if 6xⁱⁱ + x - 2 is factorable. Nosotros look for ii integers that have a product of 6(-2) = -12 and a sum of 1 (the coefficient of x). The integers iv and -3 have a product of -12 and a sum of one, so the trinomial is factorable. Nosotros now go on.

We consider all pairs of factors whose product is half-dozen. Since half dozen is positive, only positive integers need to be considered. Then possibilities are 6, i and 2, 3.
We consider all pairs of factors whose product is -2. The possibilities are 2, -1 and -2, 1. We write all possible conform ments of the factors as shown.
We select the arrangement in which the sum of products (2) and (3) yields a heart term of x.

With exercise, yous will be able to mentally check the combinations and will non need to write out all the possibilities. Paying attention to the signs in the trinomial is especially helpful for mentally eliminating possible combinations.

It is easiest to cistron a trinomial written in descending powers of the variable.

Instance 5

Factor.

a. 3 + 4xⁱⁱ + 8x
b. x - 2 + 6x^two

Solution
Rewrite each trinomial in descending powers of x and and so follow the solutions of Examples 3 and 4.

a. 4xⁱⁱ + 8x + 3
b. 6x^two + 10 - 2

As we said in Section 4.iv, if a polynomial contains a mutual monomial factor in each of its terms, nosotros should factor this monomial from the polynomial before looking for other factors.

Example 6

Factor 24ⁱⁱ - 44x - 40.

Solution
Nosotros first factor 4 from each term to go

4(6x² - 11x - ten)

We and then factor the trinomial, to obtain

4(3x + 2)(2x - 5)

Alternative METHOD OF FACTORING TRINOMIALS

If the above "trial and fault" method of factoring does not yield quick results, an culling method, which we volition at present demonstrate using the earlier case 4x² + 8x + three, may be helpful.

Nosotros know that the trinomial is factorable considering we found two numbers whose product is 12 and whose sum is eight. Those numbers are 2 and 6. We now proceed and utilize these numbers to rewrite 8x every bit 2x + 6x.

Nosotros now factor the kickoff two terms, 4*2 + 2x and the last two terms, 6x + 3.

A common factor, 2x + 1, is in each term, so we can factor again.

This is the same result that we obtained before.

4.7 FACTORING THE Departure OF TWO SQUARES

Some polynomials occur and so frequently that it is helpful to recognize these special forms, which in tum enables us to directly write their factored form. Notice that

In this section nosotros are interested in viewing this relationship from right to left, from the polynomial a² - b² to its factored form (a + b)(a - b).

The difference of ii squares, a² - b^two, equals the product of the sum a + b and the difference a - b.

Example 1

a. x2 - 9 = x2 - 32
= (x + 3)(x - 3)
b. x2 - 16 = x2 - 42
= (x + iv)(x - 4)

Since

(3x)(3x) = 9xⁱⁱ

we can view a binomial such every bit 9x² - 4 equally (3x)² - ii² and use the above method to factor.

Example 2

a. 9x^two - four = (3x)² - two²
= (3x + 2)(3x - 2)
b. 4yⁱⁱ - 25x^two = (2y)ⁱⁱ - (5x)^two = (2y + 5x)(2y - 5x)

Equally earlier, we always factor out a mutual monomial starting time whenever possible.

Instance three

a. x^three - x⁵ = x^three(l - x²)
= x³(1 + x)(l - x)
b. a^two10²y - 16y = y(a²x² - 16)
= y[(ax)² - 4²]
= y(ax - 4 )(ax + 4)

iv.eight EQUATIONS INVOLVING PARENTHESES

Often we must solve equations in which the variable occurs within parentheses. We can solve these equations in the usual manner subsequently we have simplified them by applying the distributive law to remove the parentheses.

Example 1

Solve 4(v - y) + 3(2y - 1) = 3.

Solution
We first utilise the distributive law to get

twenty - 4y + 6y - 3 = 3

Now combining like terms and solving for y yields

2y + 17 = 3

2y = -14

y=-l

The same method tin can be applied to equations involving binomial products.

Example two

Solve (x + v)(x + 3) - x = ten² + 1.

Solution
First, we utilise the FOIL method to remove parentheses and obtain

xⁱⁱ + 8x + 15 - x = x^two + ane

Now, combining like terms and solving for x yields

x² + 7x + 15 = ten² + 1

7x = -14

x = -two

four.9 Word PROBLEMS INVOLVING NUMBERS

Parentheses are useful in representing products in which the variable is contained in one or more terms in whatever gene.

Example 1

One integer is three more another. If x represents the smaller integer, represent in terms of x

a. The larger integer.
b. Five times the smaller integer.
c. Five times the larger integer.

Solution
a. x + 3
b. 5x
c. 5(x + three)

Allow united states of america say we know the sum of ii numbers is 10. If we represent one number by x, then the 2d number must be x - x every bit suggested by the post-obit tabular array.

In general, if we know the sum of two numbers is 5 and 10 represents one number, the other number must be S - x.

Instance 2

The sum of ii integers is 13. If x represents the smaller integer, correspond in terms of X

a. The larger integer.
b. Five times the smaller integer.
c. 5 times the larger integer.

Solution
a. 13 - x
b. 5x
c. 5(xiii - x)

The next instance concerns the notion of consecutive integers that was consid- ered in Department 3.8.

Example 3

The difference of the squares of two consecutive odd integers is 24. If x represents the smaller integer, represent in terms of 10

a. The larger integer
b. The square of the smaller integer
c. The square of the larger integer.

Solution

a. x + 2
b. x²
c. (x + two)ⁱⁱ

Sometimes, the mathematical models (equations) for word problems involve parentheses. Nosotros can use the approach outlined on page 115 to obtain the equation. And then, we proceed to solve the equation by first writing equivalently the equation without parentheses.

Case four

One integer is v more than a second integer. Three times the smaller integer plus twice the larger equals 45. Find the integers.

Solution

Steps i-ii
Kickoff, we write what we desire to discover (the integers) as word phrases. Then, we stand for the integers in terms of a variable.
The smaller integer: x
The larger integer: x + 5

Step 3
A sketch is not applicable.

Stride iv
At present, nosotros write an equation that represents the condition in the problem and get

3x + 2(ten + 5) = 45

Step 5
Applying the distributive law to remove parentheses yields

Step 6
The integers are vii and vii + 5 or 12.

4.ten APPLICATIONS

In this section, we will examine several applications of discussion problems that lead to equations that involve parentheses. Once again, we will follow the six steps out- lined on page 115 when we solve the problems.

COIN PROBLEMS

The basic idea of problems involving coins (or bills) is that the value of a number of coins of the same denomination is equal to the product of the value of a single coin and the full number of coins.

A table similar the i shown in the next example is helpful in solving coin issues.

Example 1

A collection of coins consisting of dimes and quarters has a value of $5.lxxx. At that place are 16 more dimes than quarters. How many dimes and quarters are in the col- lection?

Solution

Steps 1-ii We first write what nosotros desire to find as word phrases. And then, nosotros represent each phrase in terms of a variable.
The number of quarters: x
The number of dimes: x + 16

Stride 3 Next, we make a table showing the number of coins and their value.

Step 4 Now we tin can write an equation.

Step 5 Solving the equation yields

Step 6 In that location are 12 quarters and 12 + sixteen or 28 dimes in the drove.

INTEREST PROBLEMS

The basic idea of solving involvement bug is that the corporeality of involvement i earned in one year at simple involvement equals the product of the charge per unit of involvement r and the corporeality of money p invested (i = r * p). For example, $chiliad invested for one year at 9% yields i = (0.09)(one thousand) = $90.

A table like the ane shown in the next example is helpful in solving involvement problems.

Example 2

Two investments produce an annual interest of $320. $1000 more is invested at eleven% than at 10%. How much is invested at each rate?

Solution

Steps 1-ii We start write what we want to find every bit word phrases. Then, we represent each phrase in terms of a variable.
Corporeality invested at 10%: ten
Amount invested at xi%: x + 100

Step three Next, we make a table showing the amount of money invested, the rates of interest, and the amounts of interest.

Stride 4 At present, we tin write an equation relating the interest from each in- vestment and the total interest received.

Pace 5 To solve for x, first multiply each fellow member by 100 to obtain

Step 6 $1000 is invested at 10%; $yard + $1000, or $2000, is invested at 11%.

MIXTURE Problems

The basic idea of solving mixture problems is that the corporeality (or value) of the substances being mixed must equal the corporeality (or value) of the concluding mixture.

A table similar the ones shown in the following examples is helpful in solving mixture problems.

Example 3

How much candy worth 80c a kilogram (kg) must a grocer alloy with 60 kg of candy worth $1 a kilogram to make a mixture worth 900 a kilogram?

Solution

Steps 1-2 We first write what we want to find as a give-and-take phrase. Then, we represent the phrase in terms of a variable.
Kilograms of 80c processed: x

Step 3 Next, we make a table showing the types of candy, the amount of each, and the total values of each.

Step 4 We can now write an equation.

Pace 5 Solving the equation yields

Step 6 The grocer should employ 60 kg of the 800 processed.

Another type of mixture problem is i that involves the mixture of the two liquids.

Example iv

How many quarts of a 20% solution of acid should be added to 10 quarts of a xxx% solution of acrid to obtain a 25% solution?

Solution

Steps one-ii We first write what we want to observe as a word phrase. Then, nosotros represent the phrase in terms of a variable.

Number of quarts of 20% solution to be added: ten

Pace 3 Next, we make a table or drawing showing the percent of each solu- tion, the corporeality of each solution, and the amount of pure acid in each solution.

Step four We can now write an equation relating the amounts of pure acid before and after combining the solutions.

Step 5 To solve for 10, first multiply each member by 100 to obtain

20x + thirty(x) = 25(x + 10)
20x + 300 = 25x + 250
50 = 5x
10 = x

Step 6 Add 10 quarts of twenty% solution to produce the desired solution.

Chapter SUMMARY

Algebraic expressions containing parentheses tin can be written without parentheses past applying the distributive law in the form
a(b + c) = ab + ac

A polynomial that contains a monomial factor common to all terms in the polynomial tin can exist written as the product of the mutual factor and another polynomial by applying the distributive law in the form
ab + ac = a(b + c)
The distributive law can exist used to multiply binomials; the FOIL method suggests the four products involved.
Given a trinomial of the course x^two + Bx + C, if there are ii numbers, a and b, whose product is C and whose sum is B, then
x² + Bx + C = (x + a)(x + b)
otherwise, the trinomial is not factorable.
A trinomial of the grade Axⁱⁱ + Bx + C is factorable if there are ii numbers whose product is A * C and whose sum is B.
The difference of squares
a² - b² = (a + b)(a - b)
Equations involving parentheses can be solved in the usual manner after the equation has been rewritten equivalently without parentheses.