Transcript Intro to Polynomials

Ch. 5 Polynomials,
Polynomial Functions, &
Factoring
5.1 Intro to Polynomials

Polynomials
Assume: a, b, c, d, & e are constants
1. ax3 + bx2 + cx + d
(1 variable)
2. ax3y2 + bx2y2 + cxy3 + d (2 variables)
3. xy2z4 – 2xyz + 6x2 + 5
(3 variables)

Degree of Polynomial
1. In: ax3 + bx2 + cx + d
degree = 3
2. In: ax3y2 + bx2y2 + cxy3 + d
degree = 3 + 2 = 5
3. In: xy2z4 – 2xyz + 6x2 + 5
degree = 1 + 2 + 4 = 7
Adding & Subtracting Polynomials

(-7x3 + 4x2 + 3) + (4x3 + 6x2 – 13)
= -7x3 + 4x2 + 3 + 4x3 + 6x2 – 13
-x3 + 10x2 -10

(14x3 – 5x2 + x – 9) – (4x3 – 3x2 – 7x + 1)
= 14x3 – 5x2 + x – 9 – 4x3 + 3x2 + 7x – 1
= 18x3 – 2x2 + 8x – 10
5.2 Multiplication of Polynomials

Multiplying Monomials


(6x5y7)(-3x2y4)
= -18x7y11
Monomials x Polynomials

2x4(2x5 – 3x2 + 4)
= 2x4(2x5) – (2x4)(3x2) + (2x4)(4)
= 4x9 – 6x6 + 8x4

Binomial x Polynomial

(3x + 2)(2x2 – 2x + 1)
= 3x(2x2 – 2x + 1) + 2(2x2 – 2x + 1)
= 6x3 – 6x2 + 3x + 4x2 – 4x + 2
= 6x3 – 2x2 – x + 2

Binomial x Polynomial

(4xy2 + 2y)(3xy4 – 2xy2 + y)
3xy4 – 2xy2
4xy2
6xy5 – 4xy3
12x2y6 – 8x2y4 + 4xy3
12x2y6 – 8x2y4 + 6xy5
+ y
+ 2y
+ 2y2
+ 2y2

Square of a Binomial


(A + B)2 =
=
=
(A + B)2 =
(A
A2
A2
A2
+
+
+
+
B)(A + B)
AB + AB + B2
2AB + B2
2AB + B2
Geometric Interpretation of a Square
A2
AB
A
AB
B2
B
A+B
A
B

Square of A Binomial

(A - B)2 = (A - B)(A - B)
= A2 - AB - AB + B2
= A2 - 2AB + B2
(A - B)2 = A2 - 2AB + B2

(x/2 – 4y3)2
= (x/2)2 – 2(x/2)(4y3) + (4y3)2
= x2/4 – 4xy3 + 16y6

Sum / Difference of 2 Terms

(A + B)(A – B)
= A2 – AB + BA - B2
2
2
= A – B
(A + B)(A – B) = A2 – B2

Using Special Products

(7x + 5 + 4y)(7x + 5 – 4y)
= ((7x + 5) + 4y)((7x + 5) – 4y)
= (7x + 5)2 – (4y)2
= (49x2 + 70x + 25) – (16y2)
= 49x2 + 70x + 25 – 16y2
Multiplication of Polynomial Functions
f(x)·g(x)

(f· g)(x) 

Given: f(x) = x – 3
g(x) = x – 7
Find:
a) (f · g)(x)
b) (f · g)(2)
a) (f · g)(x) =
=
b) (f · g)(2) =
=
(x – 3)(x – 7)
x2 – 10x + 21
4 – 20 + 21
5
5.3 Greatest Common Factors and
Factoring by Grouping

Multiplying Polynomials


→
21x3 + 28x2
Factoring


7x2(3x + 4)
21x3 + 28x2
→ 7x2(3x + 4)
Greatest Common Factor (GCF)
Expression with largest coefficient and highest
degree that divides into each term
 12x4y3 + 6x2y2 – 3x3y
GCF = 3x2y

Your Turn

Factor
1. 16x2y3 – 24x3y4
2. -12x3y4 – 4x4y3 - 2x3y2
Solution

Factor
1. 16x2y3 – 24x3y4
= 8x2y3(2 – 3xy)
2. -12x3y4 – 4x4y3 - 2x3y2
= -2x3y2(6y2 + 2xy + 1)

GC Binomial Factor



3(x – 4) + 7a(x – 4)
= (x – 4)(3 + 7a)
7x(a + b) – (a + b)
= (a + b)(7x – 1)
Factoring by Grouping



3x2 + 12x – 2xy – 8y
= (3x2 + 12x) – (2xy + 8y)
= 3x(x + 4) – 2y(x + 4)
= (x + 4)(3x – 2y)
4x2 + 20x – 3xy – 15y
3x2 – 8y + 12x – 2xy
(Hint: rearrange and
group terms)
Your Turn

Factor
1. -4x3 + 32x2 – 20x

-4x(x2 – 8x + 5)
2. 3x(x + y) – (x + y)

(x + y)(3x – 1)
3. x3 – 3x2 + 4x - 12 (Hint: rearrange and group)

x3 + 4x – 3x2 – 12
x(x2 + 4) -3(x2 + 4)
(x2 + 4)(x – 3)
5.4 Factoring
Trinomials

Question:


How do you find a way out of a maze?
Answer:
By Trial and Error
 By always keeping your right hand on the wall


Trinomials with Leading Coefficient of 1
Recall:
(x + 3)(x + 4) = x2 + 4x + 3x + 12
= x2 + 7x + 12
 Given: x2 + bx + c
Factor the expression.
 Solution:
(x + c1)(x + c2) = x2 + c1x + c2x + c1c2
= x2 + (c1 + c2)x + c1c2
So, c1c2 = c
and 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 Turn


Factor: x2 + 9x + 18
 (x + ?)(x + ?)
(2)(9) = 18; (2 + 9) == 11
(3)(6) = 18; (3 + 6) = 9
(x + 3)(x + 9)
Factor: x2 – 11x + 24

(x - ?)(x - ?)
(-2)(-12) = 24; (-2 -12) = -24
(-3)(-8) = 24; (-3 – 8) = -11
(x – 3)(x – 8)
Your Turn

Factor: x2 -2x - 24
 (x - ?)(x + ?)
(-2)(12) = -24; (-2 + 12) = 10
(-3)(8) = -24;
(-6)(4) = -24;
(x – 6)(x + 4)
(-3 + 8) = 5
(-6 + 4) = -2

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 –B3
Thus: 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 Strategy


Factor out GCF (negative coefficient)
Number of terms in polynomial

Binomials




Trinomials




A2 + 2AB + B2 = (A + B)2
A2 – 2AB + B2 = (A – B)2
Trial & Error
4 or more


A2 – B2 = (A + B)(A – B)
A3 + B3 = (A + B)(A2 – AB + B2)
A3 – B3 = (A – B)(A2 + AB + B2)
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
 Highest degree is 2

(where a ≠ 0)

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: 1 ?2: -12
?1: 2 ?2: -6


(2x + 3)(x – 4) = 0
2x + 3 = 0
x = -3/2
x–4=0
x=4
?1: -3 ?2: 4
?1: 3 ?2: -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
1
-12
-15
2
3
4
-14
-11
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) = 0
x=5
(5, 0)
y = x2 – 10x + 25