THE DISTRIBUTIVE LAW

If we desire to main point a amount by one more number, one of two people we can multiply every term the the sum by the number prior to we include or we can very first add the terms and also then multiply. Because that example,

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In either situation the an outcome is the same.

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This property, which we very first introduced in ar 1.8, is referred to as the distributive law. In symbols,

a(b + c) = abdominal muscle + ac or (b + c)a = ba + ca

By using the distributive law to algebraic expression containing parentheses, we can attain equivalent expressions without parentheses.

Our first example requires the product of a monomial and also binomial.

Example 1 create 2x(x - 3) without parentheses.

Solution

We think that 2x(x - 3) together 2x and then apply the distributive regulation to obtain

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The above method works equally too with the product of a monomial and also trinomial.

Example 2 compose - y(y2 + 3y - 4) without parentheses.

Solution

Applying the distributive property yields

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When simplifying expressions including parentheses, we very first remove the parentheses and then incorporate like terms.

Example 3 simplify a(3 - a) - 2(a + a2).

We start by removed parentheses come obtain

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Now, combining prefer terms yields a - 3a2.

We deserve to use the distributive residential property to rewrite expression in i beg your pardon the coefficient of an expression in bracket is +1 or - 1.

Example 4 write each expression without parentheses.a. +(3a - 2b)b. -(2a - 3b)

Solution

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Notice the in example 4b, the authorize of every term is adjusted when the expression is created without parentheses. This is the same an outcome that we would certainly have obtained if we offered the steps that we introduced in ar 2.5 to leveling expressions.

FACTORING MONOMIALS indigenous POLYNOMIALS

From the symmetric building of equality, we know that if

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

Thus, if there is a monomial factor common to all terms in a polynomial, we can write the polynomial as the product the the common factor and another polynomial. Because that instance, since each term in x2 + 3x consists of x as a factor, we can write the expression as the product x(x + 3). Rewriting a polynomial in this way is called factoring, and also the number x is stated to it is in factored "from" or "out of" the polynomial x2 + 3x.

To aspect a monomial indigenous a polynomial:Write a collection of parentheses came before by the monomial common to each term in the polynomial.Divide the monomial variable into each term in the polynomial and also write the quotient in the parentheses.Generally, us can discover the usual monomial variable by inspection.

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

We can inspect that us factored appropriately by multiplying the factors and verifyingthat the product is the initial polynomial. Using example 1, we get

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If the common monomial is hard to find, we deserve to write every term in prime factored type and keep in mind the typical factors.

Example 2 variable 4x3 - 6x2 + 2x.

equipment We deserve to write

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We currently see the 2x is a common monomial variable to all 3 terms. Then we element 2x out of the polynomial, and also write 2x()

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

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and compose the quotients inside the parentheses come get

2x(2x2 - 3x + 1)

We can examine our price in instance 2 by multiply the components to obtain

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In this book, we will restrict the typical factors to monomials consist of of numerical coefficients that are integers and to integral powers of the variables. The an option of authorize for the monomial factor is a matter of convenience. Thus,

-3x2 - 6x

can it is in factored either together

-3x(x + 2) or together 3x(-x - 2)

The an initial form is usually an ext convenient.

Example 3Factor out the common monomial, consisting of -1.

a. - 3x2 - 3 xyb. -x3 - x2 + x equipment

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Sometimes it is practically to create formulas in factored form.

Example 4 a. A = p + PRT = P(1 + RT) b. S = 4kR2 - 4kr2 = 4k(R2 - r2)

4.3BINOMIAL assets I

We have the right to use the distributive legislation to multiply two binomials. Although over there is small need to multiply binomials in arithmetic as shown in the example below, the distributive law also applies to expressions containing variables.

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We will now apply the above procedure for an expression include variables.

Example 1

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

Solution First, use the distributive building to obtain

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Now, integrate like state to attain x2 + x - 6

With practice, friend will have the ability to mentally include the second and 3rd products. Theabove procedure is sometimes called the foil method. F, O, I, and L stand for: 1.The product the the first terms.2.The product of the external terms.3.The product that the inner terms.4.The product the the last terms.

The FOIL method can also be supplied to square binomials.

Example 2

Write (x + 3)2 without parentheses.Solution

First, rewrite (x + 3)2 together (x + 3)(x + 3). Next, apply the FOIL technique to get

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Combining choose terms yieldsx2 + 6x + 9

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

Example 3

create 3x(x - 2)(x + 3) without parentheses.Solution First, multiply the binomials to obtain3x(x2 + 3x - 2x - 6) = 3x(x2 + x - 6)

Now, use the distributive legislation to get 3x(x2 + x - 6) = 3x3 + 3x2 - 18x

Common Errors

Notice in instance 2

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Similarly,

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In general,

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4.4FACTORING TRINOMIALS i

In section 4.3, we saw exactly how to find the product of two binomials. Now we will reverse this process. That is, provided the product of 2 binomials, we will find the binomial factors. The procedure involved is another example the factoring. As before,we will certainly only think about factors in i beg your pardon the terms have integral numerical coefficients. Such components do not always exist, however we will study the situations where they do.

Consider the adhering to product.

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Notice that the first term in the trinomial, x2, is product (1); the last term in thetrinomial, 12, is product and also the center term in the trinomial, 7x, is the sum of commodities (2) and also (3).In general,

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We use this equation (from best to left) come factor any trinomial of the kind x2 + Bx + C. We uncover two numbers whose product is C and also whose amount is B.

Example 1 variable x2 + 7x + 12.Solution we look for 2 integers whose product is 12 and also whose sum is 7. Think about the following pairs of components whose product is 12.

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We see that the only pair of determinants whose product is 12 and also whose amount is 7 is 3 and also 4. Thus,

x2 + 7x + 12 = (x + 3)(x + 4)

Note that as soon as all terms of a trinomial are positive, we require only think about pairs of positive factors because we are searching for a pair of components whose product and also sum are positive. That is, the factored hatchet of

x2 + 7x + 12would it is in of the type

( + )( + )

When the very first and third terms the a trinomial room positive yet the middle term is negative, we require only think about pairs of negative factors due to the fact that we are looking for a pair of determinants whose product is positive but whose amount is negative. That is,the factored type of

x2 - 5x + 6

would be of the form

(-)(-)

Example 2 factor x2 - 5x + 6.

Solution due to the fact that the third term is positive and also the middle term is negative, we discover two negative integers who product is 6 and also whose sum is -5. We list the possibilities.

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We check out that the only pair of factors whose product is 6 and also whose sum is -5 is -3 and also -2. Thus,

x2 - 5x + 6 = (x - 3)(x - 2)

When the first term the a trinomial is positive and also the 3rd term is negative,the indications in the factored form are opposite. That is, the factored type of

x2 - x - 12

would be of the type

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

Example 3

Factor x2 - x - 12.

Solution we must uncover two integers who product is -12 and also whose amount is -1. We list the possibilities.

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We check out that the just pair of factors whose product is -12 and also whose amount is -1 is -4 and 3. Thus,

x2 - x - 12 = (x - 4)(x + 3)

It is less complicated to aspect a trinomial totally if any type of monimial factor typical to each term the the trinomial is factored first. For example, we can aspect

12x2 + 36x + 24

as

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A monomial deserve to then it is in factored from this binomial factors. However, first factoring the typical factor 12 from the initial expression yields

12(x2 + 3x + 2)

Factoring again, we have actually

12(* + 2)(x + 1)

which is said to be in totally factored form. In such cases, the is not vital to variable the numerical aspect itself, the is, we carry out not create 12 together 2 * 2 * 3.

instance 4

element 3x2 + 12x + 12 completely.

SolutionFirst we factor out the 3 native the trinomial to gain

3(x2 + 4x + 4)

Now, we aspect the trinomial and also obtain

3(x + 2)(x + 2)

The approaches we have developed are additionally valid because that a trinomial such as x2 + 5xy + 6y2.

Example 5Factor x2 + 5xy + 6y2.

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

x2 + 5xy + 6y2 = (x + 3y)(x + 2y)

once factoring, the is best to create the trinomial in descending powers of x. If the coefficient the the x2-term is negative, variable out a an unfavorable before proceeding.

Example 6

Factor 8 + 2x - x2.

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

-x2 + 2x + 8

Now, us can element out the -1 come obtain

-(x2 - 2x - 8)

Finally, we element the trinomial come yield

-(x- 4)(x + 2)

Sometimes, trinomials space not factorable.

Example 7

Factor x2 + 5x + 12.

Solution we look for two integers who product is 12 and whose amount is 5. Native the table in example 1 on page 149, we see that there is no pair of determinants whose product is 12 and whose sum is 5. In this case, the trinomial is not factorable.

Skill at factoring is typically the result of substantial practice. If possible, execute the factoring process mentally, writing your prize directly. Girlfriend can check the outcomes of a administrate by multiply the binomial factors and also verifying that the product is same to the offered trinomial.

4.5BINOMIAL commodities II

In this section, we usage the procedure developed in ar 4.3 to multiply binomial factors whose first-degree terms have numerical coefficients various other than 1 or - 1.

Example 1

Write as a polynomial.

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

Solution

We an initial apply the FOIL technique and then integrate like terms.

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As before, if we have actually a squared binomial, we an initial rewrite it together a product, then apply the silver paper method.

Example 2

a. (3x + 2)2 = (3x + 2)(3x + 2) = 9x2 + 6x + 6x + 4 = 9x2 + 12x + 4

b. (2x - y)2 = (2x - y)(2x - y) = 4x2 - 2xy - 2xy + y2 - 4x2 - 4xy + y2

As you may have actually seen in section 4.3, the product of 2 bionimals may have actually no first-degree hatchet in the answer.

Example 3

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

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

When a monomial factor and also two binomial determinants are gift multiplied, the iseasiest to multiply the binomials first.

Example 4

Write 3x(2x - l)(x + 2) together a polynomial.

Solution We first multiply the binomials to get3x(2x2 + 4x - x - 2) = 3x(2x2 + 3x - 2)Now multiplying by the monomial yields3x(2x2) + 3x(3x) + 3x(-2) = 6x3 + 9x2 - 6x

4.6FACTORING TRINOMIALS II

In section 4.4 us factored trinomials of the kind x2 + Bx + C wherein the second-degree term had actually a coefficient the 1. Currently we want to extend our factoring techniquesto trinomials that the type Ax2 + Bx + C, where the second-degree term has actually acoefficient other than 1 or -1.

First, we think about a check to recognize if a trinomial is factorable. A trinomial ofthe form Ax2 + Bx + C is factorable if we can find two integers who product isA * C and also whose sum is B.

Example 1

Determine if 4x2 + 8x + 3 is factorable.

Solution We examine to view if there space two integers whose product is (4)(3) = 12 and also whosesum is 8 (the coefficient that x). Consider the adhering to possibilities.

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Since the factors 6 and 2 have a amount of 8, the value of B in the trinomialAx2 + Bx + C, the trinomial is factorable.

Example 2

The trinomial 4x2 - 5x + 3 is no factorable, due to the fact that the above table mirrors thatthere is no pair of factors whose product is 12 and whose sum is -5. The check tosee if the trinomial is factorable have the right to usually be done mentally.

Once we have identified that a trinomial the the form Ax2 + Bx + C is fac-torable, we continue to uncover a pair of factors whose product is A, a pair that factorswhose product is C, and an plan that yields the suitable middle term. Weillustrate by examples.

Example 3

Factor 4x2 + 8x + 3.

Solution Above, we determined that this polynomial is factorable. We currently proceed.

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1. We take into consideration all pairs of determinants whose product is 4. Since 4 is positive, just positive integers should be considered. The possibilities space 4, 1 and 2, 2.2. We think about all pairs of factors whose product is 3. Since the center term is positive, take into consideration positive pairs of components only. The possibilities room 3, 1. We create all possible arrangements the the factors as shown.

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3. We pick the arrangement in which the sum of commodities (2) and also (3) returns a middle term the 8x.

Now, we think about the administer of a trinomial in i m sorry the constant term is negative.

Example 4

Factor 6x2 + x - 2.

Solution First, we test to view if 6x2 + x - 2 is factorable. Us look for two integers that havea product that 6(-2) = -12 and also a sum of 1 (the coefficient the x). The integers 4 and-3 have a product of -12 and a sum of 1, for this reason the trinomial is factorable. We nowproceed.

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We think about all pairs of components whose product is 6. Due to the fact that 6 is positive, only positive integers must be considered. Climate possibilities space 6, 1 and 2, 3.We consider all bag of factors whose product is -2. The possibilities space 2, -1 and -2, 1. We compose all possible arrange ments that the factors as shown.We choose the setup in i m sorry the amount of products (2) and also (3) returns a middle term the x.

With practice, you will be able to mentally inspect the combinations and also will notneed to write out all the possibilities. Paying attention to the indicators in the trinomialis specifically helpful for mentally eliminating possible combinations.

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

Example 5

Factor.

a. 3 + 4x2 + 8x b. X - 2 + 6x2

Solution Rewrite every trinomial in descending powers of x and also then follow the options ofExamples 3 and also 4.

a. 4x2 + 8x + 3 b. 6x2 + x - 2

As we stated in section 4.4, if a polynomial consists of a usual monomial factorin each of that is terms, we should element this monomial native the polynomial beforelooking for various other factors.

Example 6

Factor 242 - 44x - 40.

Solution We an initial factor 4 from each term to get

4(6x2 - 11x - 10)

We then factor the trinomial, come obtain

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

ALTERNATIVE an approach OF FACTORING TRINOMIALS

If the above "trial and error" technique of factoring does not yield fast results, analternative method, which we will certainly now show using the previously example4x2 + 8x + 3, might be helpful.

We know that the trinomial is factorable because we uncovered two number whoseproduct is 12 and also whose sum is 8. Those numbers are 2 and 6. We currently proceedand usage these number to rewrite 8x as 2x + 6x.

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We now aspect the first two terms, 4*2 + 2x and also the last two terms, 6x + 3.A typical factor, 2x + 1, is in each term, so us can aspect again.This is the same an outcome that we acquired before.

4.7FACTORING THE difference OF two SQUARES

Some polynomials happen so typically that that is beneficial to acknowledge these specialforms, which in tum allows us to straight write their factored form. Watch that

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In this section we room interested in city hall this connection from ideal to left, indigenous polynomial a2 - b2 come its factored type (a + b)(a - b).

The difference of 2 squares, a2 - b2, amounts to the product the the amount a + b and the difference a - b.

Example 1

a. X2 - 9 = x2 - 32 = (x + 3)(x - 3) b. X2 - 16 = x2 - 42 = (x + 4)(x - 4)

Since

(3x)(3x) = 9x2

we deserve to view a binomial such as 9x2 - 4 as (3x)2 - 22 and also use the above methodto factor.

Example 2

a.9x2 - 4 = (3x)2 - 22= (3x + 2)(3x - 2)b.4y2 - 25x2 = (2y)2 - (5x)2= (2y + 5x)(2y - 5x)

As before, we always factor the end a usual monomial first whenever possible.

Example 3

a.x3 - x5 = x3(l - x2) = x3(1 + x)(l - x)b.a2x2y - 16y = y(a2x2 - 16) = y<(ax)2 - 42>= y(ax - 4 )(ax + 4)

4.8EQUATIONS involving PARENTHESES

Often we need to solve equations in i m sorry the variable occurs within parentheses. Wecan solve these equations in the usual manner after ~ we have actually simplified lock byapplying the distributive regulation to remove the parentheses.

Example 1

Solve 4(5 - y) + 3(2y - 1) = 3.

Solution We first apply the distributive law to get

20 - 4y + 6y - 3 = 3

Now combining favor terms and also solving for y yields

2y + 17 = 3

2y = -14

y=-l

The same technique can be used to equations involving binomial products.

Example 2

Solve (x + 5)(x + 3) - x = x2 + 1.

Solution First, we apply the FOIL technique to remove parentheses and also obtain

x2 + 8x + 15 - x = x2 + 1

Now, combining choose terms and also solving because that x yields

x2 + 7x + 15 = x2 + 1

7x = -14

x = -2

4.9WORD problems INVOLVING NUMBERS

Parentheses are useful in representing commodities in i beg your pardon the variable is containedin one or much more terms in any factor.

Example 1

One essence is three much more than another. If x to represent the smaller integer, representin regards to x

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

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

Let united state say we recognize the amount of 2 numbers is 10. If we stand for one number byx, then the 2nd number should be 10 - x as said by the following table.

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In general, if we understand the amount of two numbers is 5 and x to represent one number,the various other number need to be S - x.

Example 2

The sum of two integers is 13. If x to represent the smaller sized integer, stand for in termsof X

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

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

The next example concerns the id of continually integers that was consid-ered in ar 3.8.

Example 3

The difference of the squares of two consecutive weird integers is 24. If x representsthe smaller sized integer, represent in terms of x

a. The bigger integerb. The square the the smaller integer c. The square the the larger integer.

Solution

a. X + 2b. X2 c. (x + 2)2

Sometimes, the mathematical models (equations) because that word difficulties involveparentheses. We can use the strategy outlined on page 115 to attain the equation.Then, we proceed to deal with the equation by first writing equivalently the equationwithout parentheses.

Example 4

One integer is five much more than a second integer. 3 times the smaller integer plustwice the larger amounts to 45. Find the integers.

Solution

Steps 1-2 First, we create what we want to discover (the integers) as word phrases. Then, we represent the integers in terms of a variable.The smaller sized integer: x The bigger integer: x + 5

Step 3 A map out is not applicable.

Step 4 Now, we write an equation the represents the condition in the problemand get

3x + 2(x + 5) = 45

Step 5 applying the distributive legislation to eliminate parentheses yields

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Step 6 The integers space 7 and also 7 + 5 or 12.

4.10 APPLICATIONS

In this section, we will examine number of applications the word problems that command toequations the involve parentheses. When again, we will follow the six measures out-lined on page 115 when we deal with the problems.

COIN PROBLEMS

The an easy idea of problems involving coins (or bills) is that the value of a numberof coins of the exact same denomination is same to the product of the worth of a singlecoin and the total number of coins.

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A table favor the one shown in the next instance is beneficial in addressing coin problems.

Example 1

A collection of coins consisting of dimes and quarters has actually a value of $5.80. Thereare 16 much more dimes 보다 quarters. How numerous dimes and also quarters room in the col-lection?

Solution

Steps 1-2 We an initial write what we desire to discover as word phrases. Then, werepresent each expression in terms of a variable.The number of quarters: x The number of dimes: x + 16

Step 3 Next, us make a table showing the number of coins and also their value.

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Step 4 currently we have the right to write an equation.

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Step 5 solving the equation yields

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Step 6 There space 12 quarters and 12 + 16 or 28 dimes in the collection.

INTEREST PROBLEMS

The simple idea of addressing interest problems is the the amount of attention i earnedin one year at basic interest equates to the product of the price of attention r and theamount that money ns invested (i = r * p). Because that example, $1000 invested for one yearat 9% returns i = (0.09)(1000) = $90.

A table prefer the one displayed in the next example is helpful in addressing interestproblems.

Example 2

Two investments produce an yearly interest of $320. $1000 an ext is invested at11% than at 10%. Just how much is invested at each rate?

Solution

Steps 1-2 We very first write what we desire to uncover as word phrases. Then, werepresent each phrase in regards to a variable. Amount invest at 10%: x Amount invest at 11%: x + 100

Step 3 Next, us make a table mirroring the amount of money invested, therates of interest, and also the amounts of interest.

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Step 4 Now, we have the right to write an equation relating the attention from each in-vestment and also the full interest received.

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Step 5 To resolve for x, very first multiply every member by 100 to obtain

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Step 6 $1000 is invest at 10%; $1000 + $1000, or $2000, is invest at11%.

MIXTURE PROBLEMS

The basic idea of resolving mixture troubles is the the amount (or value) that thesubstances being blended must same the lot (or value) of the last mixture.

A table like the ones presented in the following examples is helpful in solvingmixture problems.

Example 3

How lot candy worth 80c a kilogram (kg) have to a grocer blend through 60 kg ofcandy worth $1 a kilogram to make a mixture precious 900 a kilogram?

Solution

Steps 1-2 We an initial write what we desire to find as a native phrase. Then, werepresent the phrase in regards to a variable.Kilograms of 80c candy: x

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

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Step 4 We can now create an equation.

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Step 5 solving the equation yields

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Step 6 The grocer should use 60 kg of the 800 candy.

Another form of mixture problem is one that requires the mixture that the two liquids.

Example 4

How many quarts that a 20% equipment of acid have to be added to 10 quarts the a 30%solution of acid to achieve a 25% solution?

Solution

Steps 1-2 We very first write what we want to uncover as a word phrase. Then, werepresent the phrase in regards to a variable.

Number that quarts that 20% solution to it is in added: x

Step 3 Next, we make a table or illustration showing the percent of every solu-tion, the amount of each solution, and also the amount of pure acid in eachsolution.

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Step 4 We can now create an equation relating the quantities of pure mountain beforeand after combine the solutions.

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Step 5 To solve for x, an initial multiply every member through 100 to obtain

20x + 30(10) = 25(x + 10)20x + 300 = 25x + 250 50 = 5x 10 = x

Step 6 add 10 quarts that 20% equipment to develop the preferred solution.

CHAPTER SUMMARY

Algebraic expression containing parentheses have the right to be composed without clip byapplying the distributive regulation in the forma(b + c) = abdominal muscle + ac

A polynomial that includes a monomial factor typical to all terms in thepolynomial deserve to be composed as the product of the usual factor and anotherpolynomial by using the distributive legislation in the formab + ac = a(b + c)

The distributive law deserve to be used to multiply binomials; the FOIL an approach suggeststhe four products involved.

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Given a trinomial the the type x2 + Bx + C, if there space two numbers, a and also b,whose product is C and also whose amount is B, then x2 + Bx + C = (x + a)(x + b) otherwise, the trinomial is not factorable.

A trinomial of the type Ax2 + Bx + C is factorable if there space two numbers whoseproduct is A * C and also whose amount is B.

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The distinction of squaresa2 - b2 = (a + b)(a - b)

Equations including parentheses have the right to be fixed in the usual method after the equationhas to be rewritten equivalently without parentheses.