Ch. 5 Polynomials, Ch. 5 Polynomials, Polynomial Polynomial

Functions, & Functions, & FactoringFactoring

5.1 Intro to Polynomials5.1 Intro to Polynomials PolynomialsPolynomials

Assume: a, b, c, d, & e are constants Assume: a, b, c, d, & e are constants1.1. axax33 + bx + bx22 + cx + d + cx + d (1 variable)(1 variable)2.2. axax33yy22 + bx + bx22yy22 + cxy + cxy33 + d + d (2 variables)(2 variables)3.3. xyxy22zz44 – 2xyz + 6x – 2xyz + 6x22 + 5 + 5 (3 variables)(3 variables)

Degree of PolynomialDegree of Polynomial 1.1. In: axIn: ax33 + bx + bx22 + cx + d + cx + d

degree = 3degree = 32.2. In: axIn: ax33yy22 + bx + bx22yy22 + cxy + cxy33 + d + d

degree = 3 + 2 = 5degree = 3 + 2 = 53.3. In: xyIn: xy22zz44 – 2xyz + 6x – 2xyz + 6x22 + 5 + 5

degree = 1 + 2 + 4 = 7degree = 1 + 2 + 4 = 7

Adding & Subtracting Adding & Subtracting PolynomialsPolynomials

(-7x(-7x33 + 4x + 4x22 + 3) + (4x + 3) + (4x33 + 6x + 6x22 – 13) – 13)= -7x= -7x33 + 4x + 4x22 + 3 + 4x + 3 + 4x33 + 6x + 6x22 – 13 – 13 -x -x33 + 10x + 10x22 -10 -10

(14x(14x33 – 5x – 5x22 + x – 9) – (4x + x – 9) – (4x33 – 3x – 3x22 – 7x + 1) – 7x + 1)= 14x= 14x33 – 5x – 5x22 + x – 9 – 4x + x – 9 – 4x33 + 3x + 3x22 + 7x – 1 + 7x – 1= 18x= 18x33 – 2x – 2x22 + 8x – 10 + 8x – 10

5.2 Multiplication of 5.2 Multiplication of PolynomialsPolynomials

Multiplying MonomialsMultiplying Monomials (6x(6x55yy77)(-3x)(-3x22yy44))= -18x= -18x77yy1111

Monomials x PolynomialsMonomials x Polynomials 2x2x44(2x(2x55 – 3x – 3x22 + 4) + 4)= 2x= 2x44(2x(2x55) – (2x) – (2x44)(3x)(3x22) + (2x) + (2x44)(4))(4)= 4x= 4x99 – 6x – 6x66 + 8x + 8x44

Binomial x PolynomialBinomial x Polynomial (3x + 2)(2x(3x + 2)(2x22 – 2x + 1) – 2x + 1)= 3x(2x= 3x(2x22 – 2x + 1) + 2(2x – 2x + 1) + 2(2x22 – 2x + 1) – 2x + 1)

= 6x= 6x33 – 6x – 6x2 2 + 3x + 4x+ 3x + 4x22 – 4x + 2 – 4x + 2= 6x= 6x33 – 2x – 2x22 – x + 2 – x + 2

Binomial x PolynomialBinomial x Polynomial (4xy(4xy2 2 + 2y)(3xy+ 2y)(3xy44 – 2xy – 2xy22 + y) + y)

3xy 3xy44 – 2xy – 2xy22 + y + y 4xy 4xy2 2 + 2y + 2y 6xy 6xy55 – 4xy – 4xy33 + 2y + 2y22

12x 12x22yy66 – 8x – 8x22yy44 + 4xy + 4xy33

12x12x22yy66 – 8x – 8x22yy44 + 6xy + 6xy5 5 + 2y+ 2y22

Square of a BinomialSquare of a Binomial (A + B)(A + B)22 = (A + B)(A + B) = (A + B)(A + B) = A = A22 + AB + AB + B + AB + AB + B22

= A = A22 + 2AB + B + 2AB + B22

(A + B)(A + B)22 = A = A22 + 2AB + B + 2AB + B22

Geometric Interpretation of a SquareGeometric Interpretation of a Square

A2

AB

AB

B2

A

B

A + B

A B

Square of A BinomialSquare of A Binomial (A - B)(A - B)22 = (A - B)(A - B) = (A - B)(A - B) = A = A22 - AB - AB + B - AB - AB + B22

= A = A22 - 2AB + B - 2AB + B22

(A - B)(A - B)22 = A = A22 - 2AB + B - 2AB + B22

(x/2 – 4y(x/2 – 4y33))2 2

= (x/2)= (x/2)22 – 2(x/2)(4y – 2(x/2)(4y33) + (4y) + (4y33))22

= x= x22/4 – 4xy/4 – 4xy33 + 16y + 16y66

Sum / Difference of 2 TermsSum / Difference of 2 Terms (A + B)(A – B)(A + B)(A – B) = A = A22 – AB + BA - B – AB + BA - B22

= A = A22 – B – B22

(A + B)(A – B) = A(A + B)(A – B) = A22 – B – B22

Using Special ProductsUsing Special Products (7x + 5 + 4y)(7x + 5 – 4y)(7x + 5 + 4y)(7x + 5 – 4y)= ((7x + 5) + 4y)((7x + 5) – 4y)= ((7x + 5) + 4y)((7x + 5) – 4y)= (7x + 5)= (7x + 5)22 – (4y) – (4y)22

= (49x= (49x22 + 70x + 25) – (16y + 70x + 25) – (16y22))= 49x= 49x22 + 70x + 25 – 16y + 70x + 25 – 16y22

(f· g)(x) (f· g)(x) f(x) f(x)··g(x)g(x) Given: f(x) = x – 3Given: f(x) = x – 3 g(x) = x – 7 g(x) = x – 7

Find: Find:

a)a) (f · g)(x) (f · g)(x)

b)b) (f · g)(2)(f · g)(2)

a) (f · g)(x) = (x – 3)(x – 7)a) (f · g)(x) = (x – 3)(x – 7) = x = x22 – 10x + 21 – 10x + 21

b) (f · g)(2) = 4 – 20 + 21b) (f · g)(2) = 4 – 20 + 21 = 5 = 5

Multiplication of Polynomial Multiplication of Polynomial FunctionsFunctions

5.3 Greatest Common 5.3 Greatest Common Factors and Factoring by Factors and Factoring by

GroupingGrouping Multiplying PolynomialsMultiplying Polynomials

7x7x22(3x + 4) → 21x(3x + 4) → 21x33 + 28x + 28x22

FactoringFactoring 21x21x33 + 28x + 28x2 2 → 7x→ 7x22(3x + 4) (3x + 4)

Greatest Common Factor (GCF)Greatest Common Factor (GCF) Expression with largest coefficient and highest Expression with largest coefficient and highest

degree that divides into each term degree that divides into each term 12x12x44yy33 + 6x + 6x22yy22 – 3x – 3x33yyGCF = 3xGCF = 3x22yy

Your TurnYour Turn

FactorFactor1.1. 16x16x22yy3 3 – 24x– 24x33yy44

2.2. -12x-12x33yy44 – 4x – 4x44yy33 - 2x - 2x33yy22

SolutionSolution

FactorFactor1.1. 16x16x22yy3 3 – 24x– 24x33yy44

= = 8x8x22yy33(2 – 3xy)(2 – 3xy)

2.2. -12x-12x33yy44 – 4x – 4x44yy33 - 2x - 2x33yy22

= -2x= -2x33yy22(6y(6y22 + 2xy + 1) + 2xy + 1)

GC Binomial FactorGC Binomial Factor 3(x – 4) + 7a(x – 4) 3(x – 4) + 7a(x – 4) = (x – 4)(3 + 7a)= (x – 4)(3 + 7a)

7x(a + b) – (a + b) 7x(a + b) – (a + b) = (a + b)(7x – 1)= (a + b)(7x – 1)

Factoring by GroupingFactoring by Grouping 3x3x22 + 12x – 2xy – 8y + 12x – 2xy – 8y= (3x= (3x22 + 12x) – (2xy + 8y) + 12x) – (2xy + 8y)= 3x(x + 4) – 2y(x + 4)= 3x(x + 4) – 2y(x + 4)= (x + 4)(3x – 2y)= (x + 4)(3x – 2y)

4x4x22 + 20x – 3xy – 15y + 20x – 3xy – 15y 3x3x22 – 8y + 12x – 2xy – 8y + 12x – 2xy (Hint: rearrange and (Hint: rearrange and

group terms)group terms)

Your TurnYour Turn FactorFactor

1.1. -4x-4x33 + 32x + 32x22 – 20x – 20x -4x(x-4x(x22 – 8x + 5) – 8x + 5)

2.2. 3x(x + y) – (x + y)3x(x + y) – (x + y) (x + y)(3x – 1)(x + y)(3x – 1)

3.3. xx33 – 3x – 3x22 + 4x - 12 (Hint: rearrange and group) + 4x - 12 (Hint: rearrange and group)1.1. xx33 + 4x – 3x + 4x – 3x2 2 – 12– 12

x(xx(x22 + 4) -3(x + 4) -3(x2 2 + 4)+ 4)(x(x22 + 4)(x – 3) + 4)(x – 3)

Question:Question: How do you find a way out of a maze?How do you find a way out of a maze?

Answer:Answer: By Trial and ErrorBy Trial and Error By always keeping your right hand on the wallBy always keeping your right hand on the wall

5.4 5.4 Factoring Factoring TrinomialsTrinomials

Trinomials with Leading Coefficient of 1 Recall:

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

Given: x2 + bx + cFactor the expression.

Solution:(x + c1)(x + c2) = x2 + c1x + c2x + c1c2

= x2 + (c1 + c2)x + c1c2

So, c1c2 = cand c1 + c2 = b

Given: x2 + 5x + 6 Solution:

(x + ?)(x + ?)(2)(3) = 6(2) + (3) = 5(x + 2)(x + 3)

Given: x2 – 14x + 24 Solution:

(x - ?)(x - ?)(-2)(-12)=24 (-3)(-8)=24 (-4)(-6)=24 (-2)+(-12)= -14 (-3)+(-8)= -11 (-4)+(-6)= -10 (x – 2)(x – 12)

Given: x2 + 7x - 60 Solution:

(x + ?)(x - ?)(-2)(30) = -60 (-3)(20) = -60 (-4)(15) = -60(-2)+(30) = 18 (-3)+(20) = 17 (-4)+(15) = 11

(-5)(12) = -60 (-5)+(12) = 7

Thus,(x – 5)(x + 12)

Your TurnYour Turn

Factor: xFactor: x2 2 + 9x + 18+ 9x + 18 (x + ?)(x + ?)(x + ?)(x + ?)

(2)(9) = 18; (2 + 9) == 11(2)(9) = 18; (2 + 9) == 11(3)(6) = 18; (3 + 6) = 9(3)(6) = 18; (3 + 6) = 9(x + 3)(x + 9)(x + 3)(x + 9)

Factor: xFactor: x22 – 11x + 24 – 11x + 24 (x - ?)(x - ?)(x - ?)(x - ?)

(-2)(-12) = 24; (-2 -12) = -24(-2)(-12) = 24; (-2 -12) = -24(-3)(-8) = 24; (-3 – 8) = -11(-3)(-8) = 24; (-3 – 8) = -11(x – 3)(x – 8)(x – 3)(x – 8)

Your TurnYour Turn

Factor: xFactor: x2 2 -2x - 24-2x - 24 (x - ?)(x + ?)(x - ?)(x + ?)

(-2)(12) = -24; (-2 + 12) = 10(-2)(12) = -24; (-2 + 12) = 10(-3)(8) = -24; (-3 + 8) = 5(-3)(8) = -24; (-3 + 8) = 5(-6)(4) = -24; (-6 + 4) = -2(-6)(4) = -24; (-6 + 4) = -2(x – 6)(x + 4)(x – 6)(x + 4)

Trinomials in 2 Variables x2 – 4xy – 21y2

= (x - ?y)(x + ?y)

(-7)(3) = -21(-7)+(3) = -4)(x – 7y)(x + 3y)

x2 – 5xy + 6y2

= (x - ?y)(x + ?y)

(-2)(-3) = 6(-2)+(-3) = -5(x – 2y)(x – 3y)

Terms with Common Factors

3x3 -15x2 – 42x= 3x(x2 -5x – 14)

(2)(-7) = -14(2)+(-7) = -5

3x(x + 2)(x – 7)

Factoring by Substitution(Middle term’s degree is ½ of first term’s)

x6 – 8x3 + 15(x3)2 – 8x3 + 15(x3 + ?)(x3 + ?)

(3)(5) = 15 (-3)(-5) = 15(3)+(5)= 8 (-3) + (-5) = -8 (x3 - 3)(x3 - 5)

Trinomial Whose Leading Coefficient Is Not 1 Given: 8x2 – 10 x – 3 Solution:

1. Try 2 first terms(8x ?)(x ?)(4x ?)(2x ?)

2. Try 2 last terms(? + 1)(? - 3)(? - 1)(? + 3)

3. Try various combinations(8x + 1)(x – 3) = 8x2 + x – 24x – 3 = 8x2 – 23x - 3(8x – 1)(x + 3) = 8x2 – x – 24x – 3 = 8x2 – 25x - 3(4x + 1)((2x – 3) = 8x2 + 2x – 12x – 3 = 8x2 – 10x - 3(4x – 1)(2x + 3) = 8x2 – 2x + 12x – 3 = 8x2 + 10x - 3

Trinomial Whose Leading Coefficient Is Not 1 Given: 5x2 – 14x + 8

Solution:

1. (5x ?)(x ?)(x ?)(5x ?)

2. (? – 1)(? – 8)(? – 2 )(? – 4)

3. (5x – 1)(x – 8) = 5x2 – x – 40x – 8 = 5x2 – 41x - 8(5x – 2)(x – 4) = 5x2 – 2x – 20x + 8 = 5x2 – 22x + 8(x – 1)(5x – 8) = 5x2 – 5x – 8x + 8 = 5x2 – 13x + 8(x – 2)(5x – 4) = 5x2 – 10x – 4x + 8 = 5x2 – 14x + 8

Sum & Difference of 2 Cubes Note: (A + B)(A2 – AB + B2)

= A(A2 – AB + B2) + B(A2 – AB + B2)= A3 – A2B + AB2 + BA2 – AB2 + B3

= A3 + B3

Thus: A3 + B3 = (A + B)(A2 – AB + B2) Note (A – B)(A2 + AB + B2)

= A(A2 + AB + B2) – B(A2 + AB + B2)= A3 + A2B + AB2 – BA2 – AB2 – B3)= A3 –B3Thus: A3 – B3 = (A – B)(A2 + AB + B2)

Sum & Difference of 2 Cubes Given: x3 + 27

Solution:= (x)3 + (3)3

= (x + 3)(x2 – 3x + 9)

Given: 1 – 27x3y3

Solution:= (1)3 – (3xy)3

= (1-3xy)(1 + 3xy + 9x2y2)

5.6 General Factoring 5.6 General Factoring StrategyStrategy

Factor out GCF (negative coefficient) Number of terms in polynomial

Binomials A2 – B2 = (A + B)(A – B) A3 + B3 = (A + B)(A2 – AB + B2) A3 – B3 = (A – B)(A2 + AB + B2)

Trinomials A2 + 2AB + B2 = (A + B)2

A2 – 2AB + B2 = (A – B)2

Trial & Error 4 or more

Try factoring by grouping Check if any factor can be factored further

Examples Given: -3x2 + 12 Solution:

= -3(x2 – 4)= -3(x + 2)(x – 2)

Given: 3x2y – 12xy – 36y Solution:

= 3y(x2 – 4x – 12)= 3y(x – ?)(x + ?)= 3y(x – 6)(x + 2)

Examples Given: 9x2 + 12x + 4y2

Solution:= (3x)2 + 2(6x) + (2y)2 = (3x + 2y)2

Given: 16a2x – 25y – 25x + 16a2y (Hint: regroup terms)

Solution:= (16a2x + 16a2y) – (25x + 25y)= 16a2(x + y) – 25(x + y)= (x + y)(16a2 – 25)= (x + y)((4a)2 – (5)2)= (x + y)(4a – 5)(4a + 5)

Examples Given: 27x3 + 8 Solution:

= (3x)3 + 23

= (3x + 2)((3x)2 – 3x · 2 + (2)2)= (3x + 2)(3x2 – 6x + 4)

(3x2 – 6x + 4) = Can this be factored further?

Your Turn

Factor x10 + 512x (Hint: 512 = 83) Solution:

x((x3)3 + 83)= x(x3 + 8)((x3)2 – x38 + 82)= x(x3 + 8)(6 – 8x3 + 64)

5.8 Polynomial Equation Applications

Quadratic Equation ax2 + bx + c = 0 (where a ≠ 0) Highest degree is 2

Solving Quadratic Equation Given: 2x2 – 5x = 12 Solution:

2x2 – 5x – 12 = 0(2x ?1)(x ?2)

Possibilities: ?1: -1 ?2: 12 ?1: -2 ?2: 6 ?1: -3 ?2: 4 ?1: 1 ?2: -12 ?1: 2 ?2: -6 ?1: 3 ?2: -4

(2x + 3)(x – 4) = 0

2x + 3 = 0 x – 4 = 0x = -3/2 x = 4

Geometric Interpretation of the Solutions to the Quadric Equations 2x2 – 5x – 12 = 0

y = 2x2 – 5x – 12

x y-2 6

-1.5 0

-1 -6

0 -12

1 -15

2 -14

3 -11

4 0

(-3/2, 0)

(4, 0)

Geometric Interpretation of the Solutions to the Quadric Equations 2x2 – 5x – 12 = 0

y = 2x2 – 5x – 12 (3/2, 0) (4, 0)

(-3/2, 0)

(4, 0)

Examples

Given: x2 + 7 = 10x – 18 Solution:

x2 – 10x + 25 = 0(x - ?) (x - ?) = 0(x – 5) (x – 5) = 0x = 5

(5, 0)

y = x2 – 10x + 25