Worksheet 17 (4.1)

Chapter 4 Polynomials

4.1 Polynomials: Sums and DifferencesSummary 1:

Warm-up 1. a) The degree of 8x2y3 is .b) The degree of 2x3 - 5x2 + 3x - 6 is .

Problems

1. Determine the degree of 5x2y3 - 3xy3 - 6x3.2. Determine the degree of 5a - 6b.

Monomials are terms that contain variables with only nonnegative integers as exponents. EX. -6y2

A polynomial is a monomial or a finite sum or difference of monomials. A binomial is a polynomial with two terms. EX. 5x - 7 A trinomial is a polynomial with three terms. EX. 2x2 - 3x + 6 The degree of a monomial is the sum of the exponents of the literal factors. EX. -12a2b4c is of degree 7 Any nonzero constant term is of degree zero. EX. -8 is of degree zero The degree of a polynomial is the degree of the term with the highest degree in the polynomial. EX. 4x3y4 - 5xy4 is of degree 7

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Summary 2:

Warm-up 2. a) Add 2x2 - 5x + 6 and 3x2 + x - 4.

(2x2 - 5x + 6) + (3x2 + x - 4) = (2x2 + ) + (-5x + ) + (6 + )

= x2 - x +

b) Find the sum: (2a - 5) + (a + 6) + (3a - 2).

(2a - 5) + (a + 6) + (3a - 2) = (2a + + ) + (-5 + + )

= a - Worksheet

17 (4.1)

Problems

3. Add 3y2 - 6y + 2 and y2 - 2y - 5.

4. Find the sum: (5b - 6) + (b - 4) + (2b + 3)

Warm-up 3. a) Subtract 8x2 + 2x - 6 from -3x2 + x - 4.

(-3x2 + x - 4) - (8x2 + 2x - 6) = -3x2 + x - 4 - - + = (-3x2 - ) + (x - ) + (-4 + )

= x2 - x +

To add polynomials, rearrange, regroup and combine similar terms.

To subtract polynomials, change the problem to addition of the opposite. (To take the opposite of a polynomial, change the signs of all the terms.

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b) Simplify: 6x - [2x - (5x + 2)] = 6x - [2x - - ] = 6x - [ x - ] = 6x + x + = x +

Problems

5. Subtract -3x2 - 5x + 2 from 2x2 + 3x - 4.

6. Simplify: -3x + [8x - (7x + 9)].

Summary 3:

Warm-up 4. a) Write a polynomial to represent the perimeter of a rectangle whose width is 2x + 3 and whose length is 5x - 4. Simplify the expression by combining similar terms.

P = 2W + 2LP = 2( ) + 2( )P = x + + x - P = x -

Worksheet 17 (4.1)

b) Find the perimeter of the rectangle in part a if the value of x is 4.

P = (4) - P = - P =

The perimeter of the rectangle is .

Problems

Polynomials occur in various geometric problems: perimeter, area, volume.

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7. Write a polynomial to represent the perimeter of a square with sides of length

2x2 - 5.

8. Find the perimeter of the square in problem # 7 if x = 3.

Worksheet 18 (4.2)

4.2 Products and Quotients of Monomials

Summary 1:

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Warm-up 1. Simplify:a) y8 y4 = e) (x5)2 =

b) x x5 = f) (4x)2 = c) 52 53 = g)

d) (23)4 = h) Problems - Simplify:

1. a5 a9 2. 7 78

3. (y8)3 4. (2y)3

5. 6. Summary 2:

Properties of Exponents If a and b are real numbers and m and n are positive integers, then:

bn bm = bn + m

(bn)m = bmn

(ab)n = an bn

; when n > m; (b 0)

; when n = m; (b 0)

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Worksheet 18 (4.2)

Warm-up 2. Find each of the following products:

a) (2x2y2)(4xy4) = (24)(x2x)(y2y4) = ( )(x2 + 1)(y2 + 4) =

b) (-5x)(3x2)(2x4) = (-532)(xx2x4) = ( )( )

=

c) (-3x2y)3 = ( )3( )3( )3

=

Problems - Find each of the following products:

7. (cd2)(-3c4d5)8. (2d)(-3d3)(½d5)

9. (2x5y2)4

Summary 3:

Warm-up 3. Find each of the following quotients:

a)

= ( )( )( )

=

To multiply monomials, multiply their coefficients and multiply the variable factors by adding the exponents of the like variables.

To divide two monomials, divide the coefficients and divide the variable factors by subtracting the exponents of the like variables.

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b) = ( )( )( )

=

Problems - Find each of the following quotients:10. 11.

Worksheet 19 (4.3)

4.3 Multiplying Polynomials

Summary 1:

Warm-up 1. a) -5x2y(2x2y - 3y2 + 5) = (-5x2y)(2x2y) - (-5x2y)(3y2) + (-5x2y)(5)

= + -

Problems - Multiply:

1. 3y2(2y2 - 5y + 9)

2. -xy3(3x2y - 7xy2 + 2)

Summary 2:

To multiply a monomial and a polynomial, multiply the monomial times each term of the polynomial by using an extension of the distributive property.

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Warm-up 2. a) (x - 5)(x + 6) = x(x + 6) - 5(x + 6) = x2 + - - 30

= x2 + - 30

b) (x + 3)(x2 - 2x + 5) = x(x2 - 2x + 5) + 3(x2 - 2x + 5) = x3 - 2x2 + 5x + - + = x3 + - +

Problems - Multiply:

3. (c - 3d)(c + 5a)

4. (a - 2)(a2 + 3a - 9)Worksheet 19 (4.3)

Summary 3:

To multiply any two polynomials, multiply each term of the first polynomial times each term of the second polynomial. Simplify by combining similar terms if needed.

When multiplying two binomials, a shortened version can be used:

1. Multiply the first terms of the two binomials. (First)2. Multiply the two outer terms and multiply the two inner

terms and combine if products are similar terms. (Outer + Inner)

3. Multiply the last terms of the two binomials. (Last) This method is sometimes called the FOIL method.

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Warm-up 3. a) (x - 3)(x + 4) = + - Step 1 Step 2 Step 3

Step 1. Multiply xxStep 2. Multiply x4 and -3x and combineStep 3. Multiply -34

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

Problems - Multiply:

5. (x + 6)(x - 7)

6. (5x - 6)(2x + 4)

Summary 4:

To square a binomial you can multiply the binomial times itself or you can use the following special patterns: (a + b)2 = a2 + 2ab + b2

(a - b)2 = a2 - 2ab + b2

The pattern follows the format of: (first term)2 ± 2(first term)(second term) + (second

term)2

If the sign between the two terms is positive, the middle sign of the answer will be positive. If the sign between the two terms is negative, the middle sign of the answer will be negative.

When two binomials differ only in the middle sign a special pattern develops: (a + b)(a - b) = a2 - b2 (The sum of the outer and inner products is 0: ab - ab = 0) When cubing a binomial the following patterns develop:

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a - b)3 = a3 - 3a2b + 3ab2 - b3

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Worksheet 19 (4.3)

Warm-up 4. Multiply using the special patterns:

a) (x - 5)2 = (x)2 - 2(x)(5) + (5)2

= - +

b) (3a + 2b)2 = (3a)2 + 2(3a)(2b) + (2b)2

= + +

c) (2x - 7)(2x + 7) = ( )2 - ( )2

= -

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

= - 3( )( ) + 3( )( ) - = - + -

e) (2x + 3)3 = ( )3 + 3( )2( ) + 3( )( )2 + ( )3

= + 3( )( ) + 3( )( ) + = + + +

Problems - Use special patterns to multiply:

7. (5x - 6)2

8. (7x - 2)(7x + 2)

9. (y + 3)3

10. (3a - 4b)3

Note: If the patterns are forgotten or are difficult for you to use,

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using the distributive property will always work.

Worksheet 20 (4.4)

4.4 Factoring: Use of the Distributive Property

Summary 1:

Warm-up 1. Classify as prime or composite:a) 92 92 = 192, 246 or 423 therefore is

.b) 47 47 = 147 therefore is .

Problems - Classify as prime or composite:

1. 33 2. 79 3. 129

A prime number is a positive integer greater than 1 that has no factors that are positive integers other than itself and 1. EX. 2, 3, 5, 7

A composite number is a positive integer greater than 1 that is not a prime number. EX. 4, 6, 8, 9 Every composite number is the product of prime numbers. EX. 6 = 23 Prime factorization form of a number is the indicated product form that contains only prime factors. EX. 63 = 337 A number is completely factored when it is in the prime factorization form.

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Warm-up 2. Factor into the product of prime numbers:

a) 42 = 2 b) 140 = 2 = 23 = 22

= 225

Problems - Factor into the product of prime numbers:

4. 92 5. 125 6. 68

Summary 2:

Worksheet 20 (4.4)

Warm-up 3. Factor completely:

a) 15x3 - 5x2 = 5x2( - )b) 3x2y3 + 12xy2 = 3xy2( + )c) 5(a - b) - x(a - b) = (a - b)( - )

Note: Warm-up c) is an example of a common binomial factor.

Problems - Factor completely:

7. 24a4b - 8a3b2

8. 18x3 + 12x2 - 24x

EX. 5x - 10y = 5(x) - 5(2y) = 5(x - 2y) This type of factoring process is referred to as factoring out the highest common monomial factor. A polynomial with integral coefficients is in completely factored form if:

1. It is expressed as a product of polynomials with integral coefficients.

2. No polynomial, other than a monomial, within the factored form can be further factored into polynomials with integral coefficients.

We use the distributive property, ab + ac = a(b + c), to factor a polynomial whose terms have common factors.

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9. y(a - 2) + 5(a - 2)

10. x2(y + 5) - 8(y + 5)

Summary 3:

Warm-up 4. Factor by grouping:a) ax + bx + ac + bc = x( + ) + c( + )

= (x + c)( + )

b) 2y - 4z + ay - 2az = 2( - ) + a( - )= ( + )( - )

c) y2 - 3y - 4y + 12 = y( - ) - 4( - ) = ( - )( - )

Note: It was necessary to factor -4 from the last two terms for the binomial factors to match.

d) ac - bd + bc - ad = ac - ad + bc - bd

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

Note: It was necessary for this polynomial to be rearranged since no common factor occurred in the two pairs as it was written.Worksheet 20 (4.4)

Problems - Factor by grouping:

To factor by grouping: 1. group the four terms in pairs. 2. factor out the common factor for each pair. 3. factor out the common binomial factor which appears in step 2.

Factoring by grouping is usually used when there are four terms.

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11. 2x - ax + 6 - 3a

12. ax + bx - ay - by

13. 5m - cn + 5n - cm

Summary 4:

Warm-up 5. a) Solve: y2 + 8y = 0

y( + ) = 0 y = 0 or + = 0 y = 0 or y= Solution Set = { 0, }

b) Solve: 4c2 = 12c 4c2 - 12c= 0

4c( - ) = 0 4c = 0 or - = 0 c = or c = Solution Set = { , }

c) Solve 5ax2 - 10cx = 0 for x: 5ax2 - 10cx = 0

5x( - ) = 0 5x = 0 or - = 0

x = or x = Solution Set = { , }

x = Worksheet

The following property allows us to use factoring as a technique for solving equations:

Let a and b be real numbers, ab = 0 if and only if a = 0 or b = 0 This property tells us that if two factors multiply to give 0, one of the factors must be equal to 0.

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20 (4.4)

Problems - Solve:

14. 2x2 - 4x = 0

15. 5m2 = 15m

16. Solve for y: 2ay2 - by = 0

Warm-up 6. Set up an equation and solve:

a) The area of a square is 5 times its perimeter. Find the length of a side of the square.

A = s2 P = 4s (Geometry formulas needed.)

A = 5(P) s2 = 5(4s) s2 = 20s

s2 - = 0 s(s - ) = 0 s = 0 or s - = 0

s = 0 or s =

The length of a side of the square is .

Problems - Set up an equation and solve:

17. The area of a circular region is numerically equal to four times the circumference of the circle. Find the length of the radius of the circle.

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Worksheet 21 (4.5)

4.5 Factoring: Difference of Two Squares and Sum or Difference of Two Cubes

Summary 1:

Warm-up 1. a) x2 - 36 = ( )2 - ( )2

= ( )( )

b) 25a2 - 49 = ( )2 - ( )2

= ( )( )

c) a4 - 16b4 = ( )2 - ( )2

= ( + )( - ) = ( + )[( )2 - ( )2] = ( )( )( )

d) (a - 1)2 - b2 = [(a - 1) + ][(a - 1) - )]= ( )( )

e) 2x2 - 18 = 2( - ) = 2[( )2 - ( )2] = 2( )( )

f) x2 + 25 =

Problems - Factor completely:

1. y2 - 64 2. 16a2 - 9

Difference of Two Squares: a2 - b2 = (a + b)(a - b)

A sum of two squares will not factor. It is said to be a prime polynomial.

a2 + b2 is a prime polynomial

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3. x4 - 81 4. (x + 2)2 - y2

5. 3x2 - 48 6. a2 + 81

Worksheet 21 (4.5)

Summary 2:

Warm-up 2. a) x3 - 8 = ( )3 - ( )3

= ( - )[( )2 + ( )( ) + ( )2] = ( )( )

b) 27x3 + 1 = ( )3 + ( )3

= ( + )[( )2 - ( )( ) + ( )2] = ( )( )

c) 125y3 - 64z3 = ( )3 - ( )3

= ( - )[( )2 + ( )( ) + ( )2] = ( )( )

Problems - Factor using the sum or difference of two cubes pattern:

7. a3 - 27

8. 8x3 + 125

9. 64x3 - 1

Summary 3:

Sum of Two Cubes: a3 + b3 = (a + b)(a2 - ab + b2)

Difference of Two Cubes: a3 - b3 = (a - b)(a2 + ab + b2)

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Warm-up 3. a) Solve: 25x2= 49 25x2 - = 0

( + )( - ) = 0 + = 0 or - = 0

x = or x = x = or x =

Solution Set = { , }

Worksheet 21 (4.5)

b) Solve: 2x2 - 18 = 0 2( - ) = 0

( - ) = 0 ( + )( - ) = 0

+ = 0 or - = 0 x = or x =

Solution Set = { , }

c) Solve: x3 - 36x = 0 x( - ) = 0

x( + )( - ) = 0 x = 0 or + = 0 or - = 0 x = 0 or x = or x =

Solution Set = { , , }

d) A rectangle is three times as long as it is wide and its area is 75 square feet. Find the length and width of the rectangle.

Let x = width and 3x = length

widthlength = Area ( )( ) = 75

Special kinds of equations can be solved using the difference of two squares factoring pattern. (The two terms must be perfect squares, and the x term must be missing.)

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x2 = 75 x2 =

x2 - = 0 (x + )(x - ) = 0

= 0 or = 0 x = or x =

3x = 3( ) =

The width is and the length is .

Note: The solution x = -5 was excluded because the width can not be a negative number.

Problems - Solve:

10. x2 - 81 = 0

Worksheet 21 (4.5)

11. 4x2 = 49

12. x3 - 100x = 0

13. The cube of a number equals four times the square of the same number. Find the number.

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Worksheet 22 (4.6)

4.6 Factoring Trinomials

Summary 1:

Trinomials of the Form x2 + bx + c

Factorable trinomials such as x2 + 5x + 6 will factor into the product of two binomials; x2 + 5x + 6 = (x + 2)(x + 3), where:

1. The first terms of the two binomials multiply to give x2, the first term of the trinomial. (xx = x2)

2. The second terms of the two binomials multiply to give 6, last term of the trinomial. (23 = 6)

3. The second terms of the two binomials add to give 5, the coefficient of the middle term of the trinomial. 2 + 3 = 5

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Warm-up 1. Factor:

Note: If the last sign of the trinomial is positive, the signs in the middle of the two factors will be alike, and will be the same as the sign in the middle of the trinomial.

a) x2 + 8x + 15 = (x + )(x + ) both signs positive alike Note: The two numbers in the blanks must multiply and give 15, and add to give 8.

b) x2 - 9x + 20 = (x - )(x - ) both signs negative alike

Note: The two numbers in the blanks must multiply and give 20, and add to give - 9. Note: If the last sign of the trinomial is negative, the signs in the middle of the two factors will be different. The sign in the middle of the trinomial will go in front of the larger of the two numbers used for the second terms.

c) x2 - x - 12 = (x + )(x - )

negative signs in front of different larger number

Note: The two numbers in the blanks must multiply and give -12, and add to give -1.

Worksheet 22 (4.6)

d) x2 + 3x - 18 = (x + )(x - ) positive signs

in front of different larger number

Note: The two numbers in the blanks must multiply and give -18 and

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add to give 3.

Problems - Factor completely:

1. x2 - 7x + 10

2. x2 + 5x - 24

3. y2 + 13y + 42

4. a2 - 5a - 14

Summary 2:

Trinomials of the Form ax2 + bx + c

Factorable trinomials such as 2x2 - x - 10 will factor into the product of two

binomials; 2x2 - x - 10 = (2x - 5)(x + 2), where:1. The first terms of the two binomials multiply to give

2x2, the first term of the trinomial. (2xx = 2x2)2. The last terms of the two binomials multiply to give

-10, the last term of the trinomial. (-52 = -10)3. The products of the inner and outer terms of the

two binomials combine to give the middle term of the trinomial.

(-5x + 2x2 = -5x + 4x = -x)

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Worksheet 22 (4.6)

Warm-up 2. Factor:

a) 3x2 - 5x - 2 = (3x )(x )

the inner and outer signs products add to different

give a negative

Note: The two numbers in the blanks must multiply and give -2, and the products of the inner and outer terms must add to give -5x.

b) 6x2 + x - 12 = (2x )(3x )

the inner and outer signs products add to different

give a positive

Note: The two numbers in the blanks must multiply and give -12, and the products of the inner and outer terms must add to give +x. If 6x and 1x had been chosen for the first terms, then the product of the inner and outer terms would not have added to give +x.

c) 10x2 - 19x + 7 = (5x )(2x )

both signs negative alike

Note: The two numbers in the blanks must multiply and give 7, and the products of the inner and outer terms must add to give -19x. If 10x and 1x had been chosen for the first terms, then the product of the inner and outer terms would not have added to give -19x.

Problems - Factor completely:

5. 4x2 - 5x - 6

6. 5x2 - 22x + 8

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7. 12x2 + 4x - 5

Worksheet 22 (4.6)

Summary 3:

Warm-up 3. Factor using the perfect square trinomial pattern:

a) x2 - 18x + 81 = (x)2 -2(x)(9) + (9)2

= ( - )2

b) 16x2 + 24x + 9 = ( )2 + 2( )( ) + ( )2

= ( + )2

Problems - Factor using the perfect square trinomial pattern:

8. a2 + 8a + 16 9. 4x2 - 20x + 25

Perfect Square Trinomials

Trinomials in the form of a2 + 2ab + b2 are called perfect square trinomials since they factor into two identical factors which can be written as the square of a binomial:

a2 + 2ab + b2 = (a + b)(a + b) = (a + b)2

a2 - 2ab + b2 = (a - b)(a - b) = (a - b)2

Look for this pattern:(lst term)2 ± 2(lst term)(2nd term) + (2nd term)2 = (lst

term ± 2nd term)2

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Note: Trial and error will also work on perfect square trinomials.

Summary 4:

Worksheet 22 (4.6)

Problems - Factor completely by any method discussed in this chapter.

10. x3 + 27

1. Factor out a common monomial (or binomial) factor by using the distributive property. (Always look for this type of factoring first.) 2. If the polynomial contains 4 terms, try factoring by grouping. 3. If the polynomial contains 2 terms, look for: a difference of two squares pattern, a difference of two cubes pattern or a sum of two cubes pattern. 4. If the polynomial contains 3 terms, look for a perfect square trinomial pattern or factor by trial and error. 5. Always check to see if any factor can be factored further.

Review of Factoring Techniques

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11. ax - bx + ay - by

12. 5x2y3 - 10x3y

13. 9x2 - 12x + 4

14. 81y2 - 25

15. 6b2 + 7b - 20

Worksheet 23 (4.7)

4.7 Equations and Problem Solving

Summary 1:

In this section we continue to use the following property to solve equations:

ab = 0 if and only if a = 0 or b = 0

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Warm-up 1. a) Solve: 10x2 - x - 3 = 0

10x2 - x - 3 = 0 (5x - )(2x + ) = 0

5x - = 0 or 2x + = 0 5x = or 2x = x = or x =

Solution Set= { , }

b) Solve: -2x2 + 6x + 20 = 0

-2x2 + 6x + 20 = 0 -2(x2 - - = 0 (x2 - - ) = 0 (x - )(x + ) = 0

x - = 0 or x + = 0 x = or x =

Solution Set = { , }

c) Solve: (x + 2)(x - 1) = 10

(x + 2)(x - 1) = 10 x2 + x - 2 = 10

x2 + x = 0 (x + )(x - ) = 0 x + = 0 or x - = 0 x = or x =

Solution Set = { , }

Note: In this problem it was necessary to simplify and set = 0 before using the factoring technique to solve.

Worksheet 23 (4.7)

Problems - Solve:

1. 12x2 + 8x - 15 = 0

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2. -2a2 - 14a + 60 = 0

3. x(x - 5) = 6

Summary 2:

Warm-up 2. a) Find two consecutive even whole numbers whose product is 120.

Let x = first consecutive even whole number x + 2 = second consecutive even whole number

x( + ) = 120 x2 + = 120

x2 + - = 0 (x + )(x - ) = 0

x + = 0 or x - = 0 x = or x =

x + 2 = + 2 =

The two consecutive even whole numbers are .

Note: The negative solution for x was excluded because the problem specified a whole number.

Worksheet

Many application problems can be solved by using factoring techniques

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23 (4.7)

Problems - Solve:

4. Find two consecutive odd whole numbers whose product is 63.

5. The length of a rectangle is 5 inches more than twice the width. If the area of the rectangle is 133 square inches, find the width and length.

Summary 3:

Pythagorean Theorem

The Pythagorean Theorem is used in application problems involving right triangles. The Pythagorean Theorem states: In any right triangle, the square of the hypotenuse (the longest side, opposite the right angle), is equal to the sum of the squares of the other two sides (called legs).

The Pythagorean Theorem is sometimes stated: a2 + b2 = c2

where a and b represent the legs and c represents the hypotenuse.

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Worksheet 23 (4.7)

Warm-up 3. a) The length of one leg of a right triangle is 2 inches more than the length of the other leg. If the length of the hypotenuse is 10 inches, find the lengths of the two legs.

Let: x = length of shorter leg (a) x + 2 = length of longer leg (b)

10= length of the hypotenuse (c)

a2 + b2 = c2

x 2 + (x + 2)2 = 102

x2 + + + = 2x2 + - = 0

2( + - ) = 0 ( + - ) = 0 (x + )(x - ) = 0

x + = 0 or x - = 0 x = or x =

x + 2 =

The two legs are and .

Problem - Solve:

6. The length of one leg of a right triangle is 4 cm more than the length of the other leg. The length of the hypotenuse is 4 cm more than the length of the longer leg. Find the length of the three sides of the triangle.

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102