1. Ch. 5 Polynomials, Polynomial Functions, & Factoring

2. 5.1 Intro to Polynomials
Polynomials Assume: a, b, c, d, & e are constants
ax3 + bx2 + cx + d (1 variable)
ax3y2 + bx2y2 + cxy3 + d (2 variables)
xy2z4 – 2xyz + 6x (3 variables)
Degree of Polynomial
In: ax3 + bx2 + cx + d degree = 3
In: ax3y2 + bx2y2 + cxy3 + d degree = = 5
In: xy2z4 – 2xyz + 6x2 + 5 degree = = 7

3. Adding & Subtracting Polynomials
(-7x3 + 4x2 + 3) + (4x3 + 6x2 – 13) = -7x3 + 4x x3 + 6x2 – x3 + 10x2 -10
(14x3 – 5x2 + x – 9) – (4x3 – 3x2 – 7x + 1) = 14x3 – 5x2 + x – 9 – 4x3 + 3x2 + 7x – 1 = 18x3 – 2x2 + 8x – 10

4. 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

5. 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

6. Binomial x Polynomial
(4xy2 + 2y)(3xy4 – 2xy2 + y) xy4 – 2xy2 + y xy2 + 2y xy5 – 4xy3 + 2y2 12x2y6 – 8x2y4 + 4xy3 12x2y6 – 8x2y4 + 6xy5 + 2y2

7. Geometric Interpretation of a Square
Square of a Binomial
(A + B)2 = (A + B)(A + B) = A2 + AB + AB + B = A2 + 2AB + B2 (A + B)2 = A2 + 2AB + B2
Geometric Interpretation of a Square A2 AB B2 A B
A + B A B

8. Square of A Binomial
(A - B)2 = (A - B)(A - B) = A2 - AB - AB + B = 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

9. Sum / Difference of 2 Terms
(A + B)(A – B) = A2 – AB + BA - B2 = A2 – B2 (A + B)(A – B) = A2 – B2
Using Special Products
(7x y)(7x + 5 – 4y) = ((7x + 5) + 4y)((7x + 5) – 4y) = (7x + 5)2 – (4y)2 = (49x2 + 70x + 25) – (16y2) = 49x2 + 70x + 25 – 16y2

10. Multiplication of Polynomial Functions
(f· g)(x)  f(x)·g(x)
Given: f(x) = x – g(x) = x – 7 Find:
(f · g)(x)
(f · g)(2)
a) (f · g)(x) = (x – 3)(x – 7) = x2 – 10x + 21
b) (f · g)(2) = 4 – = 5

11. 5.3 Greatest Common Factors and Factoring by Grouping
Multiplying Polynomials
7x2(3x + 4) → 21x3 + 28x2
Factoring
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

12. Your Turn
Factor
16x2y3 – 24x3y4
-12x3y4 – 4x4y3 - 2x3y2

13. Solution Factor 16x2y3 – 24x3y4 = 8x2y3(2 – 3xy)
-12x3y4 – 4x4y3 - 2x3y2 = -2x3y2(6y2 + 2xy + 1)

14. GC Binomial Factor Factoring by Grouping
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)

15. Your Turn Factor -4x3 + 32x2 – 20x 3x(x + y) – (x + y)
x3 – 3x2 + 4x (Hint: rearrange and group)
x3 + 4x – 3x2 – 12 x(x2 + 4) -3(x2 + 4) (x2 + 4)(x – 3)

16. 5.4 Factoring Trinomials Question: Answer:
How do you find a way out of a maze?
Answer:
By Trial and Error
By always keeping your right hand on the wall

17. Trinomials with Leading Coefficient of 1
Recall: (x + 3)(x + 4) = x2 + 4x + 3x = x2 + 7x + 12
Given: x2 + bx + c Factor the expression.
Solution: (x + c1)(x + c2) = x2 + c1x + c2x + c1c = x2 + (c1 + c2)x + c1c2 So, c1c2 = c and c1 + c2 = b

18. 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)= (-3)(-8)= (-4)(-6)=24 (-2)+(-12)= (-3)+(-8)= (-4)+(-6)= -10 (x – 2)(x – 12)

19. Given: x2 + 7x - 60
Solution: (x + ?)(x - ?) (-2)(30) = (-3)(20) = (-4)(15) = -60 (-2)+(30) = 18 (-3)+(20) = (-4)+(15) = 11 (-5)(12) = -60 (-5)+(12) = 7
Thus, (x – 5)(x + 12)

20. 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)

21. Your Turn
Factor: x2 -2x - 24
(x - ?)(x + ?) (-2)(12) = -24; ( ) = 10 (-3)(8) = -24; (-3 + 8) = 5 (-6)(4) = -24; (-6 + 4) = -2 (x – 6)(x + 4)

22. 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)

23. Terms with Common Factors
3x3 -15x2 – 42x = 3x(x2 -5x – 14) (2)(-7) = -14 (2)+(-7) = -5 3x(x + 2)(x – 7)

24. Factoring by Substitution (Middle term’s degree is ½ of first term’s)
x6 – 8x (x3)2 – 8x (x3 + ?)(x3 + ?)
(3)(5) = (-3)(-5) = 15 (3)+(5)= (-3) + (-5) = -8 (x3 - 3)(x3 - 5)

25. Trinomial Whose Leading Coefficient Is Not 1 Given: 8x2 – 10 x – 3
Solution:
Try 2 first terms (8x ?)(x ?) (4x ?)(2x ?)
Try 2 last terms (? + 1)(? - 3) (? - 1)(? + 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

26. Trinomial Whose Leading Coefficient Is Not 1
Given: x2 – 14x + 8 Solution:
(5x ?)(x ?) (x ?)(5x ?)
(? – 1)(? – 8) (? – 2 )(? – 4)
(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

27. 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)

28. Sum & Difference of 2 Cubes
Given: x Solution: = (x)3 + (3)3 = (x + 3)(x2 – 3x + 9)
Given: – 27x3y3 Solution: = (1)3 – (3xy)3 = (1-3xy)(1 + 3xy + 9x2y2)

29. 5.6 General Factoring Strategy
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

30. Examples Given: -3x2 + 12 Solution: = -3(x2 – 4) = -3(x + 2)(x – 2)
Given: x2y – 12xy – 36y
Solution: = 3y(x2 – 4x – 12) = 3y(x – ?)(x + ?) = 3y(x – 6)(x + 2)

31. 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)

32. Examples
Given: 27x3 + 8
Solution: = (3x) = (3x + 2)((3x)2 – 3x · 2 + (2)2) = (3x + 2)(3x2 – 6x + 4)
(3x2 – 6x + 4) = Can this be factored further?

33. Your Turn Factor x10 + 512x (Hint: 512 = 83)
Solution: x((x3)3 + 83) = x(x3 + 8)((x3)2 – x ) = x(x3 + 8)(6 – 8x3 + 64)

34. 5.8 Polynomial Equation Applications
Quadratic Equation
ax2 + bx + c = (where a ≠ 0)
Highest degree is 2

35. Solving Quadratic Equation
Given: x2 – 5x = 12
Solution: 2x2 – 5x – 12 = 0 (2x ?1)(x ?2)
Possibilities: ?1: -1 ?2: ?1: ?2: ?1: ?2: 4 ?1: 1 ?2: ?1: 2 ?2: ?1: 3 ?2: -4
(2x + 3)(x – 4) = 0 2x + 3 = x – 4 = 0 x = -3/ x = 4

36. Geometric Interpretation of the Solutions to the Quadric Equations
2x2 – 5x – 12 = 0 y = 2x2 – 5x – 12
(-3/2, 0)
(4, 0) x y -2 6
-1.5 -1 -6
-12 1
-15 2
-14 3
-11 4

37. 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)

38. 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