Section 0.1 Multiplying Binomials - BobPrior.combobprior.com/MAT36/Trig-Chapter 0.pdf ·...

Sec 0.1 Multiplying Binomials 1 © Robert H. Prior, 2015 Copying is prohibited.

Section 0.1 Multiplying Binomials MULTIPLYING BINOMIALS, THE FOIL METHOD The product of two binomials initially results in four products, four terms. Sometimes, two of those terms are like terms and can be combined. One technique for multiplying binomials is called the FOIL method. When multiplying two binomials, such as (a + b)(c + d), there are four pairs of terms in the following positions:

Pair of terms

Position Represented by

• a and c

First terms in each binomial When we distribute, a · c is called the First product.

F

(a + b)(c + d)

F

• a and d

the outermost, or Outer, terms in the entire product When we distribute, a · d is called the Outer product.

O

O

(a + b)(c + d)

• b and c

the innermost, or Inner, terms in the entire product When we distribute, b · c is called the Inner product.

I

I

(a + b)(c + d)

• b and d

the Last terms in each binomial When we distribute, b · d is called the Last product.

L

L

(a + b)(c + d) The FOIL method doesn’t tell us anything new; it simply gives us a way to speak about the four individual products that result. The results are the same as when we apply the distributive property. In multiplying (2x + 3)(4x – 5), we get four initial products. Based on the FOIL method, those products are: F O I L First Product Outer Product Inner Product Last Product 2x · 4x = 8x2 2x · (-5) = -10x 3 · 4x = 12x 3 · (-5) = -15

8x2 – 10x + 12x – 15 = 8x2 + 2x – 15 If the Outer and Inner products are like terms, then they can be combined, making the FOIL method more like F + (O + I) + L. When the sum of O and I are done mentally, this becomes a one-step process.


Example 1: Multiply. a) (v + 5)(v + 2) b) (2x – 3)(6x – 5) c) (w2 – 5)(2w2 + 4) Procedure: Find the product using the FOIL method. These may be done in one step. Simplify

by combining like terms. Answer: a) (v + 5)(v + 2) b) (2x – 3)(6x – 5) c) (w2 – 5)(2w2 + 4) F + (O + I) + L F + (O + I) + L F + (O + I) + L = v2 + 7v + 10 = 12x2 – 28x + 15 = 2w4 – 6w2 – 20

You Try It 1 Practice using the FOIL method along with the one-step technique to multiply. Use

Example 1 as a guide.

a) (3k – 5)(k – 4) b) (2x + 3)(4x + 1) c) (3c2 + 7)(2c2 – 1)

Caution: The FOIL method does not replace anything you have already learned about multiplying polynomials. It is a technique that applies to multiplying binomials only.

SQUARING A BINOMIAL To square a binomial, such as (2x – 5)2, we can, as an option, write it as the product of the same binomial: (2x – 5)2 = (2x – 5)(2x – 5). Using the FOIL method, the product is F O I L = 4x2 – 10x – 10x + 25 = 4x2 – 20x + 25


Because the Outer and Inner products are identical, their sum is just twice the Outer product. In other words, (2x – 5)2 = (2x – 5)(2x – 5). Using the FOIL method and doubling the Outer product, we get F 2 · O L

= 4x2 – 2(10x) + 25

= 4x2 – 20x + 25 In general, the square of a binomial fits one of these patterns:

The Square of a Binomial 1. (a + b)2 = a2 + 2ab + b2 2. (a – b)2 = a2 – 2ab + b2

Example 2: Multiply these squared binomials.

a) (x + 7)2 b) (3x – 4)2 c) (2x2 + 5)2 Procedure: Multiply directly using F + 2 · O + L

Answer: a) x2 + 14x + 49 b) 9x2 – 24x + 16 c) 4x4 + 20x2 + 25

Caution: One common mistake in using this one-step technique is forgetting to

double the Outer product, which is the middle term in the resulting trinomial.

Another common mistake is to square the first and last terms but forget to

find the Outer product. This incorrectly leads to the resulting product having no middle term.

You Try It 2 Multiply these perfect square binomials. Use Example 2 as a guide.

a) (x + 8)2 b) (w – 9)2 c) (5y + 4)2 d) (6x2 – 2)2


CONJUGATES Two binomials are conjugates if their first terms are exactly the same but their second terms are opposites, as in (a + b) and (a – b). Conjugates always come in pairs. For example, (x + 3) is not a conjugate without (x – 3). When multiplying conjugates using the FOIL method, the Outer and Inner terms are always opposites and combine to 0, so the product of a pair of conjugates is always the difference of squares: (a + b)(a – b) = a2 – b2. For example, (x – 3)(x + 3) (2x + 5)(2x – 5)

= x2 + 3x – 3x – 9 = 4x2 – 10x + 10x – 25

= x2 + 0x – 9 = 4x2 + 0x – 25

= x2 – 9 = 4x2 – 25

Example 3: Identify the conjugate of the given binomial and then multiply the pair of conjugates.

a) (2r – 3) b) (6y + 7) b) (3y2 + 5) Procedure: The product of a pair of conjugates is always the difference of squares. Answer:

a) Conjugate is (2r + 3) b) Conjugate is (6y – 7) b) Conjugate is (3y2 – 5)

(2r – 3)(2r + 3) (6y + 7)(6y – 7) (3y2 – 5)(3y2 – 5)

a2 – b2 a2 – b2 a2 – b2

= (2r)2 – (3)2 = (6y)2 – (7)2 = (3y2)2 – (5)2

= 4r2 – 9 = 36y2 – 49 = 9y4 – 25

You Try It 3 Identify the binomial conjugate of the given binomial and then multiply the pair of

conjugates. Use Example 3 as a guide.

a) (x – 8) b) (4x + 9) c) (10w2 – 2)


You Try It Answers YTI 1: a) 3k2 – 17k + 20 b) 8x2 + 14x + 3 c) 6c4 + 11c2 – 7 YTI 2: a) x2 + 16x + 64 b) w2 – 18w + 81 c) 25y2 + 40y + 16 d) 36x4 – 24x2 + 4 YTI 3: a) (x + 8); x2 – 64 b) (4x – 9); 16x2 – 81 c) (10w2 + 2); 100w4 – 4

Section 0.1 Focus Exercises Multiply and simplify. Write each answer in descending order. 1. (x + 3)(x + 8) 2. (y – 2)(y – 6) 3. (u + 4)(u – 2) 4. (w – 10)(w + 3) 5. (2n + 5)(n + 1) 6. (x + 6)(3x – 4) 7. (2r2 – 3)(5r2 – 1) 8. (3r2 + 5)(6r2 + 2) 9. (x + 7)2 10. (w – 9)2


11. (y – 3)2 12. (m + 4)2 13. (3c – 5)2 14. (4p + 3)2 15. (a2 + 10)2 16. (2x2 – 5)2 Identify the binomial conjugate of the given binomial and then multiply the pair of conjugates. 17. (x + 5) 18. (y – 6) 19. (r – 8) 20. (p + 10) 21. (2c + 9) 22. (4w – 7) 23. (x2 – 8) 24. (6y2 – 1)

Sec. 0.2 Factoring Trinomials 7 © Robert H. Prior, 2015 Copying is prohibited.

Section 0.2 Factoring Trinomials FACTORING QUADRATICS

A quadratic polynomial is in the form ax2 + bx + c. In this section we discuss a one-step method for factoring trinomials such as a) 2x2 + 7x + 6 b) 3w2 – 20w + 12 c) 6m2 + 7m – 5 = (2x + 3)(x + 2) = (w – 6)(3w – 2) = (3m + 5)(2m – 1) Because factoring is the reverse process of multiplication, we can think about reversing the steps of multiplying binomials and apply it to factoring trinomials. For example, consider multiplying (3x + 2)(x – 6). We start by finding the four FOIL products: (3x + 2)(x – 6) F O I L The four FOIL products: = 3x2 – 18x + 2x – 12

The two middle terms are like terms and can combine, giving us a trinomial: = 3x2 – 16x – 12

The reverse of this includes the following steps. Starting with a trinomial 3x2 – 16x – 12, write the middle term as two like terms. This will create a four-term polynomial which we can factor using factoring by grouping: Start with a trinomial: 3x2 – 16x – 12

Split the middle term into two terms and write the trinomial with four terms: = 3x2 – 18x + 2x – 12

Use factoring by grouping: = (3x2 – 18x) + (2x – 12) = 3x(x – 6) + 2(x – 6) and here is the factored form: = (3x + 2)(x – 6) The most important question in all of this is, how do we know what two like terms (in this case -18x and +2x) to write in place of the middle term, -16x? The answer to this question is, the Factor Game. THE FACTOR GAME The Factor Game is a game of two numbers: a Product number and a Sum number. The game is to find a factor pair of the Product number that add to get the Sum number. The factor pair that works is called the “winning combination,” or just “combination.”


Example 1: Find the winning combination of the Factor Game with the given product and sum numbers.

a) Product = +20 and sum = -9. b) Product = -30 and sum = +1 Procedure: For a), the product is positive and the sum is negative so both factors in the factor pair

must be negative. For b), the product is negative, so one factor must be positive and the other factor must be negative.

Answer: a) Factor pairs of +20: b) Factor pairs of -30: +20 Sum -30 Sum -1 -20 -21 -1 +30 +29 -2 -10 -12 -2 +15 +13 -4 -5 -9 This is it! -3 +10 +7 -5 +6 +1 This is it!

You Try It 1 Find the winning combination of the Factor Game with the given product and sum numbers. Use Example 1 as a guide.

a) product = +32 b) product = -36 c) product = -40 sum = -12 sum = +16 sum = -6 TRINOMIALS AND THE FACTOR GAME

Now let’s put the Factor Game to good use. Consider a trinomial of the form ax2 + bx + c, where a, b, and c are integers, and a > 0.

In the trinomial ax2 + bx + c, the Product number is a· c, and the Sum number is b.

Example 2: Given the trinomial, identify the Product and Sum numbers, and find the winning combination of the Factor Game, if there is one.

a) 3x2 – 14x + 8 b) 5y2 – 4y – 12 c) 3w2 + 6w – 8 Procedure: Identify the values of a, b, and c and find the product and sum numbers of the Factor

Game. Note: If a trinomial has no winning combination, then it is not factorable.

Answer:

a) Product = 3(8) = 24Sum = -14 b)

Product = 5(-12) = -60Sum = -4 c)

Product = 3(- 8) = - 24Sum = 6

Combination : -12 and -2 Combination :6 and -10 There is no combination.


You Try It 2 Given the trinomial, identify the Product and Sum numbers, and find the winning combination for the Factor Game, if there is one. Use Example 2 as a guide.

a) 4x2 + 12x + 5 Product = b) 5x2 – 13x + 6 Product = Sum = Sum = Combination: Combination: c) 6x2 – 13x – 4 Product = d) 3x2 + 4x – 4 Product = Sum = Sum = Combination: Combination: FACTORING TRINOMIALS USING THE FACTOR GAME

The whole point of the Factor Game is to be able to rewrite a trinomial into four terms. It is the winning combination of the Factor Game that tells us how to split up the trinomial’s middle term into two terms, thereby making it a four-term polynomial.

Consider factoring the trinomial 8x2 + 10x – 3. Here are the steps to factoring a trinomial using the Factor Game and factoring by grouping.

(1) Identify the Product and Sum numbers and find the combination to the Factor Game:

8x2 + 10x – 3 Product = 8(-3) = -24Sum = + 10 The combination is +12 and -2.

The two factors in the combination are the coefficients of the two new x-terms that make the trinomial into a four-term polynomial. In other words, the middle term, + 10x, can be replaced by + 12x – 2x:

8x2 + 10x – 3

(2) Write the trinomial with four terms: = 8x2 + 12x – 2x – 3

(3) Use factoring by grouping: = (8x2 + 12x) + (- 2x – 3) = 4x(2x + 3) + - 1(2x + 3) = (4x – 1)(2x + 3) We can verify that this factoring is true by multiplying the binomials together using FOIL. If the Factor Game has no winning combination, then the trinomial is not factorable, and we say that it is prime.


Example 3: Factor each trinomial by using the Factor Game to rewrite it as a four-term polynomial. Then use factoring by grouping.

a) 6p2 + 5p – 6 b) 5x2 – 16x + 12 c) 3r2 + 6r – 8 Procedure: Identify the Product and Sum numbers and play the Factor Game. Use the winning

combination to write the trinomial as a four-term polynomial, and use factoring by grouping. Verify that the result is true by using FOIL to multiply the results.

Answer:

a) 6p2 + 5p – 6 Product = 6(-6) = -36 Sum = + 5 The combination is -4 and +9.

6p2 + 5p – 6 Split the middle term into – 4p + 9p : = 6p2 – 4p + 9p – 6 Now show the groupings. (There must be a plus sign between the groupings.

= (6p2 – 4p) + (9p – 6) Factor out the common monomial factor from each group … = 2p(3p – 2) + 3(3p – 2) … and factor out (3p – 2). = (2p + 3)(3p – 2) Now verify this factoring by multiplying the binomials.

b) 5x2 – 16x + 12 Product = 5(12) = 60 Sum = - 16 The combination is -10 and -6.

5x2 – 16x + 12 Split the middle term into - 10x – 6x: = 5x2 – 10x – 6x + 12 Now show the groupings. (There must be a plus sign between the groupings. = (5x2 – 10x) + (- 6x + 12) Factor out the common monomial factor from each group. = 5x(x – 2) + - 6(x – 2) Now factor out (x – 2): = (5x – 6)(x – 2) Now verify this factoring by multiplying the binomials.

c) 3r2 + 6r – 8 Product = 3(-8) = -24 Sum = + 6 There is no combination.

Because there is no combination to the Factor Game, the trinomial is prime.


You Try It 3 Factor each trinomial by using the Factor Game to rewrite it as a four-term polynomial. Then use factoring by grouping. If the trinomial cannot be factored, write “prime.” Use Example 3 as a guide.

a) 3y2 + 11y + 6 b) 2x2 – 5x + 4 c) m2 + m – 30 d) 6p2 – p – 2

ONE-STEP TRINOMIAL FACTORING

Let’s take another look at the trinomial in Example 3a):

6p2 + 5p – 6 The winning combination is -4 and +9, so the middle term is split into - 4p + 9p F O I L

= 6p2 – 4p + 9p – 6 Notice that the F product, 6p2, and the O product, -4p, are paired up in (6p2 – 4p).

= (6p2 – 4p) + (9p – 6)

= 2p(3p – 2) + 3(3p – 2) Notice also that their greatest common factor (GCF), 2p, is the first term of the first binomial factor (in the answer).

= (2p + 3)(3p – 2) This bit of information will serve as the foundation for factoring a trinomial in one step. We must now learn how to find the other three terms of the binomial factors.

For example, factor 8x2 + 10x – 3 in just one step: A. Identify the four FOIL products within the trinomial, and play the Factor Game.

We already have the F and L products: F L

8x2 + 10x – 3


It’s just a matter of finding the O and I products from the Factor Game:

8x2 + 10x – 3 Product = 8(-3) = -24 Sum = +10 The combination is +12 and -2.

Either 12x or -2x can be the O product, but it is common to choose the larger one to be O.

Let’s choose the O product = +12x

and the lesser as the I product = -2x

The GCF of the F and O products is 4x, and that is the first term of the first binomial factor. B. Create the framework; Fill in the binomial factors; Verify the factoring

1. Because the trinomial is factorable, it will factor into two binomials, and we can write two sets of parentheses in anticipation of those binomial factors:

8x2 + 10x – 3

( )( )

2. Instead of writing the Outer and Inner products as

8x2 + 12x – 2x – 3, we write the O product above the parentheses, as shown.

8x + 10x – 32

( )( )

+ 12x

3. The first term of the first binomial is the GCF of the F and O products. The GCF of 8x2 and 12x is 4x.

Now that we have the “first of the first” we can use it and the

FOIL products to find other terms within the two binomials.

8x + 10x – 32

+12x

4x

GCF

4. We can use the first term, 4x, and the F product, 8x2, to help us find the first term of the second binomial.

8x + 10x – 32

( )( )4x 2x

F

5. Next, we find the O product. We already have the first term of the O product, 4x. To create the O product, +12x, the second term of the second binomial must be +3.

( )( )

+12x

4x + 32x

6. To find the only remaining unknown term, we can use the L product, -3. This is the product of the two constant terms. Because one of the constant terms is already known, +3, the other must be -1, and this is the value that is placed in the first binomial.

8x + 10x – 32

4x + 32x– 1

7. To complete the factoring, we must verify that it is accurate by multiplying the two binomials together. This can be done mentally or on paper using FOIL.

(4x – 1)(2x + 3) = 8x2 + 12x – 2x – 3 = 8x2 + 10x – 3


One final note about factoring trinomials:

If the two factors in the combination of the factor game are exactly the same, then the trinomial is a perfect square trinomial.

For example, for 25y2 + 20y + 4, we get Product = +100 Sum = +20 The combination is +10 and +10.

and 25y2 + 20y + 4 factors into (5y + 2)(5y + 2) which can be written as (5y + 2)2. You Try It 4 Factor each using the one-step method. Use the discussion above as a guide.

a) 10x2 – 11x + 3 b) 8m2 – 2m – 3 c) 10r2 + 7r – 6 d) x2 – 18x + 81 e) 9y2 – 6y + 1 f) 9y2 + 9y – 4 THE DIFFERENCE OF SQUARES Just as a pair of conjugates multiplies to the difference of squares, we can factor the difference of squares into a pair of conjugates. For example,

Examples: x2 – 25 4y2 – 81w2 w4 – 16

= (x + 5)(x – 5) = (2y – 9w)(2y + 9w) = (w2 + 4)(w2 – 4)

= (w2 + 4)(w + 2)(w – 2) Note: The sum of squares is not factorable; it is prime. For example, x2 + 4 is prime.


You Try It 5 Factor each binomial. Use the discussion above as a guide.

a) x2 – 100 b) 4m2 – 49 c) 25x2 + 1 d) y3 – 36y e) 3p3 – 75p f) x4 – 81

You Try It Answers: Section 0.2 YTI 1: a) -4 and -8 b) +18 and -2 c) -10 and +4 YTI 2: a) Product = +20, Sum = +12; combination is +10 and +2 b) Product = +30, Sum = -13; combination is -10 and -3 c) Product = -24, Sum = -13; There is no combination. d) Product = -12, Sum = 4; combination is -2 and +6 YTI 3: a) (3y + 2)(y + 3) b) Prime. c) (m – 5)(m + 6) d) (3p – 2)(2p + 1) YTI 4: a) (5x – 3)(2x – 1) b) (4m – 3)(2m + 1) c) (2r – 1)(5r + 6) d) (x – 9)2 e) (3y – 1)2 f) (3y – 1)(3y + 4) YTI 5: a) (x + 10)(x – 10) b) (2m – 7)(2m + 7) c) Prime d) y(y – 6)(y + 6) e) 3p(p + 5)(p – 5) f) (x2 + 9)(x + 3)(x – 3)


Section 0.2 Focus Exercises Factor each trinomial by using the Factor Game. You may use any method presented in this section. If the trinomial is not factorable, then write “prime.”

1. 4x2 + 11x + 6 2. 2w2 – 9w + 9 3. 10y2 + 9y + 2 4. p2 – 12p + 36 5. 5v2 – 6v – 8 6. 4x2 – 12x + 5 7. 4m2 + 6m + 9 8. 12k2 – 8k + 1


9. h2 – 16h + 64 10. 10x2 + 3x – 4 11. 15r2 + 4r – 3 12. 6p2 + 7p – 10 Factor each binomial. 13. x2 – 144 14. 9m2 – 64 15. 49x2 – 1 16. x2 + 100 17. 81x2 – 16 18. w3 – 25w 19. 5m3 – 45m 20. 16x4 – 1

Sec 0.3 Solving Quadratic Equations 17 © Robert H. Prior, 2015 Copying is prohibited.

Section 0.3 Solving Quadratic Equations INTRODUCTION A quadratic equation is an equation in which the degree of the polynomial is 2.

The standard form for a quadratic equation is

ax2 + bx + c = 0 or 0 = ax2 + bx + c Standard form means that one side of this equation is a polynomial in descending order and the other side is 0. Some quadratic equations are not yet written in standard form but can be rewritten into standard form. Here are some examples of quadratic equations:

- 2y2 + 6y = 0 v2 – 7v = - 10 (x + 3)(x – 2) = 14

THE ZERO PRODUCT PRINCIPLE The product of any two non-zero numbers will always be either positive or negative. The only way a product can be zero is if one of the factors is zero. This idea leads us to the Zero Product Principle.

The Zero Product Principle

If the product of two numbers is 0, then one of the numbers must be 0.

If A · B = 0, then either A = 0 or B = 0. The Zero Product Principle can be applied to any two factors that multiply to get 0. For example, if (x + 7)(x – 5) = 0, then one of the factors must be 0: either x + 7 = 0 or x – 5 = 0 In other words, either x = - 7 or x = 5 To apply the Zero Product Principle, one side of the equation must be 0 and the other side must be in a factored form. For example, we cannot apply the Zero Product Principle on x2 + x – 6 = 0 until the left side of the equation is factored:

x2 + x – 6 = 0 (x + 3)(x – 2) = 0

Either x + 3 = 0 or x – 2 = 0 Either x = -3 or x = 2 It is common to combine the solutions into a single solution set: x = -3, 2


Example 1: Solve each equation. a) �x2 + 2x – 35 = 0 b) -2r2 – 8r = 0 c) 0 = 4y2 – 13y + 3 Procedure: Factor the polynomial and apply the Zero Product Principle. Verify the answers and

write the solutions in a solution set. Answer: a) x2 + 2x – 35 = 0 Factor the left side. Use the Factor Game. (x + 7)(x – 5) = 0 Set each factor equal to 0. x + 7 = 0 or x – 5 = 0 Solve each linear equation. x = -7 or x = 5 Write the solutions in a solution set. x = -7, 5 b) -2r2 – 8r = 0 Factor the left side. Extract -2r. -2r(r + 4) = 0 Set each factor equal to 0. -2r = 0 or r + 4 = 0 Solve each linear equation. r = 0 or r = -4 A solution may be 0. r = 0, -4 c) 0 = 4y2 – 13y + 3 Factor the right side. Use the Factor Game. 0 = (4y – 1)(y – 3) Set each factor equal to 0. 4y – 1 = 0 or y – 3 = 0 Solve each linear equation.

y = 14 or y = 3 A solution may be an integer or a fraction.

y = 14 , 3

Notice that, in Example 1 c), 0 is on the left side. That is okay. The Zero Product Principle works the same way as long as 0 is on one side or the other.


You Try It 1 Solve each equation. Use Example 1 as a guide. a) m2 + 3m – 28 = 0 b) - 5x2 + 20x = 0 c) 4w2 – 5w – 6 = 0 d) 9k2 – 16 = 0

To solve a quadratic equation, the polynomial must first be set equal to 0. Consider x2 + 3x + 2 = 12. As written, the polynomial is not yet equal to 0, so we cannot yet factor. Instead we must add -12 to each side to create 0 on the right side. Then we may factor and solve: x2 + 3x + 2 = 12 First, add -12 to each side so that 0 is on the right side: -12 = -12 Now factor the polynomial: x2 + 3x – 10 = 0 (x + 5)(x – 2) = 0 Write the two linear equations and solve: x + 5 = 0 or x – 2 = 0 x = - 5 or x = 2 x = -5, 2


Example 2: Solve each equation.

a) k2 – 5k + 2 = -4 b) (y + 6)(2y – 3) = -28 c) 4x – x2 = 4 Procedure: Get 0 (zero) on one side of the equation. Factor the polynomial and apply the Zero

Product Principle. Verify the answers and write the solutions in a solution set. Answer:

a) k2 – 5k + 2 = -4 Add 4 to each side to create 0 on the right side.

k2 – 5k + 6 = 0 Factor the left side. Use the Factor Game. (k – 2)(k – 3) = 0 Set each factor equal to 0. k – 2 = 0 or k – 3 = 0 Solve each linear equation. k = 2 or k = 3 Combine the solutions. k = 2, 3 b) (y + 6)(2y – 3) = -28 First, multiply out the left side.

2y2 + 9y – 18 = -28 Add 28 to each side to create 0 on the right side.

2y2 + 9y + 10 = 0 Factor the left side. Use the Factor Game. (2y + 5)(y + 2) = 0 Set each factor equal to 0. 2y + 5 = 0 or y + 2 = 0 Solve each linear equation.

y = - 52 or y = -2 Combine the solutions.

y = - 52 , 2

c) 4x – x2 = 4 Because the x2 term on the left is negative, add x2 and -4x to each side to create 0 on the left.

0 = x2 – 4x + 4 Factor the left side. Use the Factor Game. This is a perfect square trinomial.

0 = (x – 2)2 There is only one solution. Set the single factor equal to 0 and solve. x – 2 = 0 x = 2


You Try It 2 Solve each equation. Use Example 2 as a guide. Check your answers to verify that they are solutions.

a) x2 + 5x – 8 = 28 b) 6w = 5w2 – 8 c) y2 + 2y = 8y – 9 d) (x – 6)2 = x + 6 e) (3v – 1)(v – 5) = -15 f) (x – 1)(6x + 5) = x + 3


THE SQUARE ROOT PROPERTY OF EQUATIONS

Consider the equation x2 = 25. There are two ways that we can approach this. Either technique we choose, though, should result in the same solution set. Technique 1: Solve by factoring Technique 2: Solve by taking the square root x2 = 25 Add -25 to each side

x2 – 25 = 0 Factor (x – 5)(x + 5) = 0 Solve.

x = 5, -5 Two Solutions.

x2 = 25

x2 = 25 This appears to have only one solution, x = 5. However, we know ithere are two solutions, and we can represent them as x = ± 5

This is an example of the Sqaure Root Property: If x2 = a

then x2 = a

and x = ± a

Example 3: Solve (y + 4)2 = 36 using the Square Root Property of Equations. Procedure: Take the square root of each side. The following step must include the ± symbol in

front of the evaluated square root.

Answer: (y + 4)2 = 36 Take the square root of each side.

(y + 4)2 = 36 For the next step, place ± in front of the 6 y + 4 = ±6 Set y + 4 equal to 6 and then to -6. y + 4 = 6 or y + 4 = -6 Solve each linear equation. y = 2 or y = -10 Combine the solutions. y = 2, -10

You Try It 3 Solve each equation using the Square Root Property of Equations.. Use Example

3 as a guide. Check your answers to verify that they are solutions. a) (x – 5)2 = 16 b) (2y + 3)2 = 49


You Try It Answers

YTI 1: a) m = -7, 4 b) x = 0, 4 c) w = - 34 , 2 d) k = -

43 ,

43

YTI 2: a) x = -9, 4 b) w = -4 5 , 2 c) y = 3 d) x = 3, 10

YTI 3: a) x = 1, 9 b) y = 2, -5

Section 0.3 Focus Exercises Solve each equation. Check your answers to verify that they are solutions.

1. 6x2 – 54x = 0 2. 3r2 + 12r = 0 3. p2 – p – 90 = 0 4. 2n2 – 13n + 15 = 0 5. 3p2 + p = 10 6. w2 + 2w – 4 = 59


7. -3x2 + 4x = -15 8. y2 – y = 18 – 4y 9. v(8v + 2) = 15 10. (x – 3)(x + 7) = -16 11. x2 = 81 12. w2 = 6 13. (m – 9)2 = 4 14. (4x + 1)2 = 25

Sec 0.4 Rationalizing the Denominator 25 © Robert H. Prior, 2015 Copying is prohibited.

Section 0.4 Rationalizing the Denominator RATIONALIZING THE DENOMINATOR

If a radical expression contains an irrational denominator, such as 73 ,

712 , or

205 , then it is not

considered to be in a simplified form. To simplify such a fraction, we use a process called rationalizing the denominator. To rationalize a denominator means to make it into a rational number—specifically, a whole number. To rationalize a denominator requires us to create a perfect square radicand in the denominator. We do

this by multiplying the fractions by a creative form of 1, cc .

For example, we can multiply 73 by

33 to create 9 in the denominator, and we simplify it from

there: 73 ·

33 =

219 =

213

In the example above, we say that the common multiplier is 3 , but it means that we must multiply both the numerator and the denominator by 3 .

Note: Radicals are compatible with multiplication and division, but radicals are incompatible with addition and subtraction:

1 a · b = a · b and 2 ab =

a b

Likewise, 3 a · b = a · b and 4 ab =

ab

However, a + b ≠ a + b and a + b ≠ a + b

Example 1: Rationalize the denominator and simplify each expression.

a) 36 b)

205

Procedure: Multiply by an appropriate form of 1 to create a perfect square radicand in the

denominator. Simplify if possible.

Answer: a) 36 =

36 ·

66 =

1836 =

9 · 26 =

3 26 =

22

b) 20

5 = 20

5 · 55 =

20 55 = 4 5


Sometimes we can simplify the fraction before rationalizing. For example, in 1a), we could have made

36 into a single radical:

36 and then simplify it to

12 and then

12 before rationalizing the

denominator. Sometimes the denominator can be simplified before rationalizing. For example, the 12 in the

denominator of 5

12 can first be simplified to 4 · 3 = 2 3 :

5

12 = 5

2 3 · 33 =

152 9 =

152 · 3 =

156

You Try It 1 Rationalize the denominator and simplify each expression. Use Example 1 as a

guide.

a) 72

b) 12

3

c) 38

d) 1218

WHY WE RATIONALIZE THE DENOMINATOR

We know how to rationalize the denominator, but why do we do it? To understand this, let’s consider the decimal approximation of 2 :

2 ≈ 1.4142

In trigonometry the number 12 appears quite a bit. Where is this number located on the number line?

Is it more than 1 or less than 1?


To answer these questions, we must find its decimal approximation. One way is to use long division:

12 ≈

11.4142 → 1.4142 1.000000 (shown at right)

The long division tells us 12 ≈ 0. 7071

0. 7 0 7 1.... 1.4142 1.0 0 0 0 0 0 0

– 9 8 9 9 4 1 0 0 6 0 0 – 9 8 9 9 4 1 6 0 6 0 – 1 4 1 4 2 1 9 1 8

However, the process is much simpler if we first rationalize the denominator:

12 ·

22 =

22 ≈

1.41422

This division is much simpler because the divisor is a whole number (shown at right):

This long division also tells us 2

2 ≈ 0. 7071

0. 7 0 7 1.... 2 1.4142 0 0

– 1 4 0 1 4 – 1 4 0 2 – 0 2 0

Example 2: Given 11 ≈ 3.3166, find the value of 111 .

Answer: Rationalize the denominator and then use long

division.

111

· 1111

= 11

11

So, 111 ≈ 0.3015

0. 3 0 1 5... 11 3. 3 1 6 6

– 3 3 0 1 6 – 1 1 5 6 – 5 5 1

RATIONALIZING WITH A BINOMIAL DENOMINATOR The product of a binomial radical expression and its conjugate always results in a rational number, oftentimes an integer. For example,


1. The conjugate of ( )7 + 3 is ( )7 – 3 and their product is

( )7 + 3 ( )7 – 3 = ( )7 2 – ( )3 2 = 7 – 3 = 4 2. The conjugate of ( )2 – 10 is ( )2 + 10 and their product is

( )2 + 10 ( )2 – 10 = 22 – ( )10 2 = 4 – 10 = -6

This knowledge is useful when a radical expression has a binomial in the denominator, such as 10

3 – 7 .

To rationalize the denominator, we use the conjugate of the denominator as the common multiplier. For

example, to rationalize the denominator of 10

3 – 7 , the common multiplier is 3 + 7 , so we multiply

the fraction by 3 + 73 + 7 :

10

3 – 7 · 3 + 73 + 7 =

10(3 + 7)9 – 7 =

10(3 + 7)2 = 5(3 + 7 ) = 15 + 5 7

Example 3: Rationalize the denominator and simplify 1 – 32 + 3

Procedure: Multiply by an appropriate form of 1 to create a perfect square radicand in the

denominator. The common multiplier is the conjugate of the denominator.

Answer: 1 – 32 + 3 Multiply by

2 – 32 – 3 .

= 1 – 32 + 3 ·

2 – 32 – 3 Multiply the fractions. Multiplying the

denominators creates the difference of squares.

= (1 – 3) · (2 – 3)

(2)2 – ( 3)2 Multiply out the numerator and simplify the denominator.

= 2 – 3 – 2 3 + 9

4 – 3 Simplify the denominator as well as the numerator radicals.

= 2 – 3 3 + 3

1 The numerator does not simplify any further.

= 5 – 3 3 Whew!


You Try It 2 Rationalize the denominator and simplify each expression. Use Example 3 as a

guide.

a) -8

1 – 5

b) 2 – 8

2 – 3

You Try It Answers

You Try It 1: a) 142 b) 4 3 c)

64 d) 2 2

You Try It 2: a) 2 + 2 5 b) -2 2 – 6


Section 0.4 Focus Exercises For each, identify the conjugate of the binomial and then find the product of the pair of conjugates. 1. 2 – 3 2. 1 + 5 3. 6 – 2 4. 3 2 + 7 5. 10 – 2 2 6. 2 5 + 3 5 For each, rationalize the denominator and simplify the expression.

7. 12

3 8. 85

9. 156 10.

58

11. 7

18 12. 2050


Given the approximate value of the radical, find the value of the fraction by first rationalizing the denominator and then using long division. Round your result to three or four decimal places. 13. 3 ≈ 1.732; use this to find the 14. 5 ≈ 2.236; use this to find the

approximate value of 13 approximate value of

15 .

15. 6 ≈ 2.4495; use this to find the 16. 10 ≈ 3.1623; use this to find the

approximate value of 16 . approximate value of

110 .


For each, rationalize the denominator and simplify the expression.

17. 5

4 + 11

18. 1

1 + 2

19. 5

2 – 5

20. 2 – 8

2 – 3

Sec 0.5 Complex Fractions 33 © Robert H. Prior, 2015 Copying is prohibited.

Section 0.5 Complex Fractions A complex fraction is any fraction in which the numerator or the denominator, or both, contain one or more fractions. Examples of complex fractions include

5 78

23

6 1

52

x2y

3y2

23 +

56

54 –

12

1x –

32x

1 + 5x2

SIMPLIFYING COMPLEX FRACTIONS, METHOD 1 If a complex fraction contains no addition or subtraction, then it can be simplified by first writing the fraction as division.

For example, 23

54

can be rewritten as 23 ÷

54 . This can be simplified by inverting the second

fraction and multiplying: 23 ·

45 =

815 , which cannot simplify any further.

Example 1: Simplify each complex fraction by first rewriting it using the division symbol.

a) 5 78

b) w

xw

c) 23

6 d)

x2y

3y2

Procedure: First rewrite each using the division symbol and then invert and multiply.

Answer: a) 5 78

= 5 ÷ 78 =

51 ·

87 =

407

b) w

xw

= w ÷ xw =

w1 ·

wx =

w2x

c) 23

6 = 23 ÷

61 =

23 ·

16 =

218 =

19

d)

x2y

3y2

= x

2y ÷ 3y2 =

x2y ·

y23 =

xy26y =

xy6


You Try It 1 Simplify each complex fraction by first rewriting it using the division symbol. Use Example 1 as a guide.

a) 23

75

b) 10

54

c)

xy

3xy2

SIMPLIFYING COMPLEX FRACTIONS, METHOD 2 A second method of simplifying complex fractions is to clear all denominators directly, while in its

complex form. We do so by multiplying the whole fraction by a carefully chosen value of 1, such as 77

, xx , or

3w2

3w2 .

For example, in the complex fraction

45 32

, we can clear the denominators, 2 and 5, by using the

common multiplier of 10, multiplying the whole fraction by 1010 . (10 is the least common denominator

for 45 and 32 ). In this case, 1010 is better written as

101 101

so that we can more easily multiply:

45 32

·

101 101

= 45 ·

101

32 ·

101

=

405 302

= 8

15

Note: We will use LCD to abbreviate least common denominator. Let’s put Method 2 into practice.


Example 2: Simplify each complex fraction by using Method 2.

a) 5

78

b)

78

34

c)

3x

2x2

d)

x2y

3y2

Procedure: First recognize the least common denominator (LCD), then use it as the common

multiplier. In part a), write the numerator as a fraction, 51 . Follow each step

carefully.

Answer: a) LCD = 8:

51

78

·

81

81

=

51 ·

81

78 ·

81

=

51 ·

81

71 ·

11

= 407

b) LCD = 8:

78

34

·

81

81

= 78 ·

81

34 ·

81

= 71 ·

11

31 ·

21

= 76

c) LCD = x2:

3x

2x2

·

x21

x21

= 3x ·

x21

2x2 ·

x21

= 31 ·

x1

21 ·

11

= 3x2

d) LCD = 2y2:

x2y

3y2

·

2y21

2y21

=

x2y ·

2y21

3y2 ·

2y21

= x1 ·

y1

31 ·

21

= xy6

You Try It 2 Simplify each complex fraction by using Method 2. Use Example 2 as a guide.

a)

58 32

b)

49 83

c)

4y2

w2 8y3w


WHEN A COMPLEX FRACTION IS, WELL ... MORE COMPLEX If a complex fraction contains addition or subtraction, we cannot use Method 1 without first creating a single fraction in both the numerator and denominator. As an alternative, we can use Method 2 and multiply the complex fraction by the LCD of all of the smaller fractions within.

For example, the complex fraction

23 +

56

54 –

12

has four denominators within it: 3, 6, 4, and 2. The LCD

is 12, so the common multiplier for the entire fraction is 12, and we use it to clear all of the denominators within the complex fraction.

Multiply the whole fraction by 1212 , better written as

121 121

, as shown here:

= 23 +

56

54 –

12

· 12

1 121

Multiply

121 to both the numerator and the denominator.

Distribute 121 to each term in the numerator

and to each term in the denominator.

= 23 · 12

1 + 56 · 12

1 54 · 12

1 – 12 · 12

1 Simplify each product of fractions within by dividing out common factors.

= 8 + 1015 – 6 Again, simplify in the numerator and in the denominator.

= 189 = 2 Simplify this fraction.

Your work might look a little more like this:

12

1

12

1

·

2

3

5

6

5

4

1

2

+2

3

5

6

12

1

5

4

1

2

12

1

·+

–

12

12

1·

·

·

1

1

1

1

4

3

2

6

8 10+

15 6–

18

92= = =

... or you might distribute and cancel mentally as you go, such as ...

23 +

56

54 –

12

· 12

1 121

= 8 + 1015 – 6 =

189 = 2


Example 3: Simplify the complex fraction by using Method 2, described above.

4x –

32x

1 + 5x2

Procedure: Make every term a fraction, if it isn’t already. That will help in the multiplication

process. Next, find the least common denominator and use it to create a form of 1.

Answer: First, the common denominator is 2x2. Let’s go right to the distribution step where

each small fraction is multiplied by the 2x21 .

4x · 2x2

1 – 32x ·

2x21

11 · 2x2

1 + 5x2 · 2x2

1 =

41 · 2x

1 – 31 · x1

11 · 2x2

1 + 51 · 21

= 8x – 3x

2x2 + 10 = 5x

2x2 + 10

You Try It 3 Simplify each complex fraction using Method 2. Use Example 3 as a guide.

a) 38 +

16

2 – 14

b) v4 –

1v

12 –

1v

You Try It Answers

You Try It 1: a) 1021 b) 8 c)

y3

You Try It 2: a) 5

12 b) 16 c)

3y2w

You Try It 3: a) 1342 b)

v + 22 or

v2 + 1


Section 0.5 Focus Exercises

Simplify each complex fraction using any method.

1.

53 74

2.

109 2512

3.

411 8 4.

12 65

5.

3mp

5m2p

6.

u3v 6u

Simplify each complex fraction using Method 2.

7. 1 –

52

1 + 23

8.

512 +

13

23 –

14

9. 23 –

12

1 – 16

10. 1 +

2y

y3 +

23

11. 16 –

12w

13w –

1w2

12. 1 –

1y2

1y +

1y2

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