1. 12-1 12-2 12-3 12-4 12-5 12-6 12-7 12-8 Chapter Review
Algebra Chapter 12
Chapter Review

2. 12.1
12-1 Graphing y-k = a(x-h)2
Objectives: Understand that the graph of a function with equation y-k = a(x-h)2 is a parabola with vertex (h,k)
Vertex form of an equation of a parabola:
y-k = a(x-h)2

3. Activity 1: Set graphing calculator window:
-15 ≤ x ≤ 15 ; ≤ y ≤ 10
a. On your calculator, Graph the equation y+3 = (x-5)2
b. Graph the parabola onto paper.
c. Label the vertex with its coordinate.
d. Draw and label the axis of symmetry.
e. How do the coordinates of the vertex relate to the equation?
f. Write the equation in standard form.
12.1
x=5
(5,-3)
The x-coordinate of the vertex is subtracted from x and the y-coordinate of the vertex is subtracted from y
y = x2 – 10x+22 the vertex is not obvious from this equation

4. Find the vertex of this parabola.
12.1
y-6 = 3(x+4)2
Find the vertex of this parabola.
Remember: in the equation y-k = a(x-h)2 The vertex is (h,k)
In this equation the vertex will be (-4, 6)

5. 2) Write an equation in vertex form for the graph.
12.1
2) Write an equation in vertex form for the graph.
Step 1: get the vertex coordinates: vertex: (-3,2)
Step 2: write the vertex form of the equation.
y-2 = a(x+3) 2
Step 3: find another point with whole number coordinates in the graph. Ex: (-2,4)
Step 4: substitute the point from step 3 in for x & y and solve for a.
4-2 = a(-2+3) 2
2 = a
Step 5: write the equation with a
Answer: y-2 = 2(x+3)2

6. 3) Write an equation in vertex form for the graph.
12.1
3) Write an equation in vertex form for the graph.
Vertex: (-1,-1)
1 point with whole coordinates: (1,-3)
y - k = a(x-h)2
y + 1 = a(x + 1)2
= a (1 + 1)2
-2 = a(4)
-.5 = a
y + 1 = -.5(x + 1)2

7. 12-2 Completing the Square
12.2
12-2 Completing the Square
Objectives:
Use completing the square as a method for converting a quadratic expression in standard form to vertex form.

8. How to “complete the square”
Definition
Completing the square means finding a number “c” that when used with x2+bx will allow you to factor the created trinomial into (x-r)2 or (x+r)2
How to “complete the square”
1. Start with x2 + bx
2. Divide “b” by 2 and square it – this will be used as your 3rd term in your trinomial You get x2 + bx + (b/2)2.
12.2

9. 12.2
Complete the square, then use the formulas from chapter 11 to factor the trinomial you created.
x2 + 8x 2) x2 - 10x
3) m2 – 3m 4) y2 + 7y
+ 16 = (x+4)(x+4)
+ 25
= (x-5)(x-5)
8/2 = 4; 42 = 16 = c
+ 2.25
= (x-1.5)(x-1.5)
= (x+3.5)(x+3.5)
Remember: Divide “b” by 2 and square it – this will be used as your 3rd term in your perfect square trinomial You get x2 + bx + (b/2)2 = (x + b/2)2

10. x2 - 4x - 5 = y x2 - 4x - 5 = 0 replace y with 0
Example of using completing the square to find the roots of an equation (no quadratic formula)
When finding the roots of an equation – these are the x-intercepts so y must equal 0.
x2 - 4x - 5 = y x2 - 4x - 5 = 0 replace y with 0
x2 - 4x = add 5 to both sides; isolate “x2+bx”
x2 - 4x + 4 = completes the square – add to both sides
(x - 2)2 = 9 factor left, simplify right
x - 2 = 3 or x - 2 = -3 sq root both sides x = 5 or x = -1
12.2

11. Find the x-intercepts by completing the square, identify the vertex.
5) x2 + 8x + 12 = y
12.2
x2 + 8x + 12 – 12 = 0 – 12 remove “c” using property of equality
x2 + 8x = -12 simplify
x2 + 8x = complete the square; use prop or equality
x2 + 8x = simplify
(x+4)2 = factor left side
x + 4 = +/ take sq root of both sides
x = -6 or x = solve
The vertex is ½ way between the roots – symmetry. What is the axis of symmetry?
x = -4
Substitute -4 for x, solve for y to find the vertex.
The vertex is (-4,-4)

12. Find the roots to the equation by completing the square, identify the vertex.
6) m2 – 3m – = y
12.2
x = 5.5 or -2.5
The vertex is (1.5,-16)

13. 7) Write in vertex form & find the vertex:
x2 - 10x + 2 = y
x2 - 10x = y – 2
x2 - 10x + 25 = y + 23
(x - 5)2 = y + 23
Vertex = (5, - 23)

14. Find the roots of the equation by completing the square, identify the vertex.
8) x2 - 10x + 1 = y
x = 2sqr(6) + 5 or -2 sqrt(6) + 5
The vertex is (5,-24)

15. 12-3 The Factored Form of a Quadratic Function
12.3
12-3 The Factored Form of a Quadratic Function
Objective: Learn the advantages of the factored form of a quadratic expression in interpreting the graph of a quadratic function.

16. 12.3
Definition:
Factored form – the equation y = ax2 + bx+ c is in factored form when it is written as y = a(x – r1)(x – r2) where r1 & r2 are the x-intercepts of the parabola.

17. Graph each equation and find the x-intercepts: a) y = (x+4)(x-5)
1) Example
Graph each equation and find the x-intercepts:
a) y = (x+4)(x-5)
b) y = 3(x+1)(x+2)
c) y = (x-1)(x-2)
d) y = x(x-4)
e) What is the relationship between the equation and the x-intercepts?
12.3
(-4,0) & (5,0)
(-1,0) & (-2,0)
(1,0) & (2,0)
(0,0) & (4,0)
In y = a(x – r1)(x – r2)
r1 & r2 are the x-intercepts of the parabola

18. Graph the following equations on your calculator: Equ 1: Equ 2: Equ 3:
2) Example:
Graph the following equations on your calculator:
Equ 1: Equ 2: Equ 3:
y+4 = (x+4)2 y =x2 + 8x y = (x+6)(x+2)
a) What’s different about each graph?
b) What’s the same?
c) What information does each equation give us?
12.3
Nothing
They are the same parabola.
Equ 1 gives the vertex which is (-4, -4).
Equ 2 is standard form - simplified gives the degree: AND y-intercept is 12.
Equ 3 gives the x-intercepts (-6,0) and (-2,0).

19. What’s the coefficient of each x?
3) Without graphing, what are the x-intercepts of the parabola for y = (x-5)(x+7)?
What’s the coefficient of each x?
4) What is the value of y at the x-intercepts? Without graphing, what are the x-intercepts of the parabola for y = (2x+6)(2x-1). X-intercepts can be found using 0 = (2x+6)(2x-1).
12.3
(5,0) and (-7,0) 1
2x+6 = 0 OR 2x-1 = 0 (2-8 Zero product property)
2x = OR 2x = 1
x = OR x = .5 these are the x-intercepts
This is how to find x-intercepts when the coefficient of x is not 1.
“Solve the equation” means when y = 0, find x-intercepts.

20. 6) Find the vertex of the parabola: y = (2x + 4)(x+6)
5) How does the x-coordinate of the vertex relate to the x-intercepts? Remember that parabolas are symmetric…
6) Find the vertex of the parabola: y = (2x + 4)(x+6)
12.3
The x-coordinate of the vertex is the mean
of the 2 x-intercept x-coordinates.
X-intercepts are: (-2,0) and (-6,0)
X-coordinate of the vertex is (-4,?)
Substitute x back into the equation… y = (2(-4)+4)((-4)+6)
y = -8 solve for y  vertex is (-4,-8)
These 3 points are enough to sketch the graph

21. Find intercepts: 0=2x-1 OR 0=x-5.5
7) Find the x-intercepts and the vertex of the parabola: y = 2(2x - 1)(x – 5.5). And
y = (2x - 1)(x – 5.5).
a) What’s the same? b) What’s different?
12.3
Find intercepts: 0=2x-1 OR 0=x-5.5
Both x-intercepts: (½, 0) and (5.5, 0)
Find vertex x-coordinate: (½ + 5.5) / 2 = 3
y = 2(2(3) - 1)(3-5.5) = 25
Eq1 Vertex is (3, -25)
y = (2(3) - 1)(3-5.5) = -12.5
Eq2 Vertex is (3 ,-12.5 )

22. Summary:
Because the expression is quadratic, a not = 0, so either x- r1 =0 or x- r2 =0. Thus either x= r1 or x= r2. All quadratics of the form y=(x- r1 )(x- r2 ) have the same x-intercepts regardless of the value of a.
The x value of the vertex will be the same but the y-value will be different as the value of a changes.
12.3

23. 12-4 Factoring x2 + bx + c Objective: Factor trinomials where a = 1.
12.4
12-4 Factoring x2 + bx + c
Objective: Factor trinomials where a = 1.
Determine whether a quadratic polynomial can be factored over the integers.
Represent quadratic expressions and their factorizations with areas.

24. Steps to factor a trinomial:
12.4
Definition:
Factored form – the equation y = ax2 + bx+ c is in factored form when it is written as y = a(x – r1)(x – r2) where r1 & r2 are the x-intercepts of the parabola.
Steps to factor a trinomial:
If “a” is 1: Step 1: identify factor pairs of “c”
Step 2: of all the factors pairs find the pair that adds to “b”

25. 12.4
A quadratic trinomial is given. Factor completely over the integers and then check by multiplying.
1) x2 + 50x + 96
STEP 1:
Factor pairs
for “c”
1 · 96
2 · 48
3 · 32
4 · 24
6 · 16
8 · 12
STEP 2:
Sum of Factor pairs
to add to “b”
1+ 96 = 97
= 50
= 35
= 28
= 22
= 20
STEP 3: Use that information to create the binomials.
Answer: (x + 2)(x + 48)
Check : FOIL
= x2 + 2x+ 48x+ 96
= x2 + 50x + 96
= 50

26. Factor completely over the integers and then check by multiplying.
2) y2 – 3y – 54
12.4
STEP 2:
Sum of Factor pairs
to add to “b”
= 53
= -53
= 25
= -25
= 15
= -15
= 3
= -3
STEP 1:
Factor pairs
of “c”
-1 · 54
1 · -54
-2 · 27
2 · -27
-3 · 18
3 · -18
-6 · 9
6 · -9
= -3
Answer: (x-9) (x+6)

27. Factor completely over the integers and then check by multiplying.
3) z2 – 9z ) a2 - a - 42
5) c2 – ) d3 – 16d
12.4
Answer: (x-5) (x-4)
Answer: (x-7) (x+6)
Answer: (x+4) (x-4)
Answer: d (d+4) (d-4)

28. 7) Explain why c2 + 16 is not factorable over the integers.
12.4
There are no 2 numbers that both multiply to 16 and add to 0.
Not all trinomials are factorable – some need to use the quadratic formula.
x = the discriminant is negative…
Therefore there are no real-number solutions.

29. 12-5 Factoring ax2 + bx + c Objective: Factor trinomials where a ≠ 1.
12.5
12-5 Factoring ax2 + bx + c
Objective: Factor trinomials where a ≠ 1.

30. METHOD 1: Factor the trinomial into 2 binomials: 21n2 + 76n + 20
12.5
METHOD 1: Factor the trinomial into 2 binomials: 21n n
STEP 1: Factor “a”
(1n + ?) (21n + ?)
(3n + ?) (7n + ?)
STEP 2: Find factor pairs of “c”
1 · 20
2 · 10
4 · 5
STEP 3: Look for “b”; Find needed factor pair through “FOIL”
(1n + 1) (21n + 20): b= (1n + 2) (21n + 10): b = 52
(1n + 20) (21n + 1): b= (1n + 10) (21n + 2): b = 212
(3n + 1) (7n + 20): b= (3n + 2) (7n + 10): b=44
(3n + 20) (7n + 1): b= (3n + 10) (7n + 2): b=76
Solution found….
Answer: (3n + 10)(7n + 2)

31. METHOD 2: Factor the trinomial into 2 binomials: 21n2 + 76n + 20
1) Multiply (a)(c) = 2) Find factors that add to “b” = 3) Make a table where “b” is split: 21n2 +70n +6n +20 4) Factor across and down. Use those to make the binomials.
21 · 20 = 420
70 · 6 = 420
= 76 7n 2 3n 10
(3n+ 10)(7n + 2)

32. METHOD3: 1st Always factor out GCF (if there is one) Next Steps: 1
METHOD3: 1st Always factor out GCF (if there is one) Next Steps: 1. Take x2 coeff. Away and multiply it by the constant (c) 2. Factor using “OLD SCHOOL” 3. Put x2 coefficient back into each binomial x term ( ) 4. Factor out GCF from each ( ) if possible and leave remaining binomials

33. ExA 10x2 -29x +21 1. x2 -29x +210 mult const: 21 by x2 coefficient 2
ExA 10x2 -29x x2 -29x +210 mult const: 21 by x2 coefficient 2. find factors of 210 (both neg. b/c middle term is neg.) which add to -29: (x-14) (x-15) 3. (10x- 14) (10x- 15) mult 10 for both x’s 4. (5x- 7) (2x- 3) factor out GCF for both factors and keep the remaining factors

34. Ex B 4x2 +25x -21 1. x2 +25x -84 mult. const: -21 by x2 coeff. :4 2
Ex B 4x2 +25x x2 +25x -84 mult. const: -21 by x2 coeff.:4 2. find factors of -84 (both neg. b/c middle term is neg.) which add to 25: (x+28) (x-3) 3. (4x+28) (4x- 3) mult 4 for both x’s 4. (x +7) (4x- 3) factor out GCF for 1st factor

35. METHOD 4 – AC Method
5x2 +32x +12
5x2 +30x +2x +12 Split x term to create common factors for terms
5x(x+6)2(x+6) Factor 1st two and 2nd two terms
(5x+2)(x+6) Factor out common factors

36. 2) Factor 2x2 + 5x + 3 Answer: (x + 1)(2x + 3) STEP 1: Factor “a”
12.5
STEP 1: Factor “a”
(1x + ?) (2x + ?)
STEP 2: Find factor pairs of “c”
1 · 3
3 · 1
Answer: (x + 1)(2x + 3)

37. 3) Factor the trinomial 2n2 – 3n – 20 Answer: (n - 4)(2n + 5)
b. 6y2 - 29y - 5
12.5
Answer: (n - 4)(2n + 5)
Answer: (y - 5)(6y + 1)

38. 4) Solve the equation a. 2n2 - 3n - 20 = 0 Answer: (n - 4)(2n + 5)
b. 6y2 - 29y - 5 = 0
12.5
Answer: (n - 4)(2n + 5)
n = 4 or -2.5
Answer: (y - 5)(6y + 1)
y = 5 or -1/6

39. Combining Concepts
a) Find the expressions for the area and the perimeter of the rectangle.
b) If the area = the perimeter – what is x?
A = -x2 – 10x; P = 20
-x2 – 10x = 20
x2 + 10x = -20 make “a” positive 1 x
10-x
x2 + 10x + 25 = complete the square.
(x+5)2 = 5 simplify and write in binomial form.
x+5 = take square root of both sides
x = subtract 5 from both sides
This rectangle cannot exist – distances are not negative.

40. 12-6 Which Quadratic Expressions Are Factorable?
12.6
12-6 Which Quadratic Expressions Are Factorable?
Objectives: Use the discriminant b2 – 4ac and the fact that when a, b, and c are integers, the discriminant must be a perfect square or 0 in order for ax2 + bx + c to be factorable over the integers.
Example 6

41. b2 – 4ac is a perfect square.
Theorem:
When a, b and c are integers, with a ≠ 0, either all three of the following conditions hold, or none of these hold.
b2 – 4ac is a perfect square.
ax2 + bx + c is factorable over the set of polynomials with integer coefficients.
The solutions to ax2 + bx + c = 0 are rational numbers.
12.6

42. b. Verify your answer by solving -2 + 5t + 8t2 = 0.
12.6 1)
Is t + 8t2 factorable into polynomials with integer coefficients?
b. Verify your answer by solving t + 8t2 = 0.
Calculate the discriminant: b2 - 4ac = (8)(-2) = 89
Since the discriminant is 89 and is NOT a perfect square or 0 – then this is not factorable over integers.
Use the quadratic formula to solve:
≈ OR

43. 2) Solve 2x2 – 3x = 2 by any method.
12.6
2x2 - 3x - 2 = rewrite as a quadratic
Discriminant is (-3)2 - 4(2)(-2) = 9+16 = 25
so it is factorable……
Method 1: Factor
(2x+1)(x-2) this is factorable
x = -½ or 2
Method 2: Use the quadratic formula:
x = -½ or 2

44. 2a) Solve by completing the square. n2 - 9n + 20 = y
12.6
n2 - 9n + 20 = x intercepts are where y = 0
n2 - 9n = get rid of “c”
n2 - 9n = complete the square
(n - 4.5)2 = complete the square
take square root of both sides
n = or
n = 5 or 4
n = ±.5

45. 3) Solve 3x2 + 8x = -4 by any method.
12.6
3x2 + 8x + 4 = rewrite as a quadratic
Discriminant is (8)2 - 4(3)(4) = 64 – 48 = 16
so it is factorable……
Method 1: Factor
(3x+2)(x+2) this is factorable
x = -⅔ or -2
Method 2: Use the quadratic formula:
x = -⅔ or -2

46. Discriminant is (5)2 - 4(3)(7) = 25 – 84 = negative
4) Solve 3x2 + 5x + 7 = 0
5) Solve 4x2 – 21 = 0
12.6
Discriminant is (5)2 - 4(3)(7) = 25 – 84 = negative
There is no real number solution.
Difference of 2 “squares”… Factor over real numbers
(2x ) ·
(2x ) = 0
0 = (2x ) OR 0 =
(2x )
x = ±

47. 12.6
Solve
6) 4z3 + 28z2 + 48z = 0
4z (z2 + 7z + 12) = rewrite as a quadratic
4z (z + 3)(z + 4) = factor the quadratic
4z = Zero Product Property OR
z + 3 = 0
z + 4 = 0
z = 0 OR -3 OR solve

48. Review
7) Factor 3a2 + 2ab - 8b2
12.6
(3a + ?b) (a - ?b) OR (3a - ?b) (a + ?b)
Remember Factors of -8: -1 & 8, 1 & -8, -2 & 4, 2 & -4
(3a + ?b) (a - ?b) FOIL must add to +2 but multiply to -8
Extreme Factors : -1 & 8, 1 & -8, probably wont work –
they’re too different 24 vs 1 to add to 2 so
try -2 & 4, 2 & -4 first
(3a - 4b) (a + 2b)

49. 12.7
12-7
Objectives : Make a connection between the x-intercepts of the graphs and factors of quadratic expressions to polynomials of degree higher than 2. .

50. a) f(x) = x-3 b) g(x) = x2 - 4 c) h(x) = (x+1)(x)(x-2)
12.7
Activity 1: Graph the following on your calculator, give their x-intercepts and degree of the polynomial.
a) f(x) = x b) g(x) = x c) h(x) = (x+1)(x)(x-2)
Make a conjecture about the factors of any degree polynomial and the x-intercepts of its graph.
x-int = 3
Deg = 1
x-int = 2, -2
Deg = 2
x-int = -1, 0, 2
Deg = 3
g(x) = x2 - 4
h(x) = (x+1)(x)(x-2)
f(x)=x-3
The factors of the polynomial identify the x-intercepts.
The number of x-intercepts will equal the polynomial’s degree

51. There are 4 intercepts and the polynomial will have a degree of 4.
12.7
2) Test your conjecture from Activity 1 with: j(x) = (x+1)(x-1)(x+2)(x-2) - use your graphing calculator to help you sketch the graph into your notes.
x-intercepts are: -2,-1,1,2
There are 4 intercepts and the polynomial will have a degree of 4.
j(x) = (x+1)(x-1)(x+2)(x-2)

52. Let r be a real number and P(x) be a polynomial in x.
12.7
Factor Theorem
Let r be a real number and P(x) be a polynomial in x.
1) If x-r is a factor of P(x), then P(r) = 0; that is, r is an x-intercept of the graph of P.
2) If P(r) = 0, then x-r is a factor of P(x).

53. 12.7
3) A polynomial Q(x) has x-intercepts of -8, -2, 4, and 20. Give a possible equation for this function and tell the degree the function will have.
4) Give a possible equation for the graph. What is the degree of the equation and the y-int?
Answer: (x+8)(x+2)(x-4)(x-8)(x-20) This function will have a degree of 5.
Answer: y = (x+4)(x+¾)(x-¼)(x-3) The y-int is (0,1).This function will have a degree of 4.

54. 7) How does the y-int relate to the equation?
12.7
5) Rewrite h(x) = (x+1)(x)(x-2) from Activity 1 into standard form. Identify the y-intercept before looking back at your graph.
6) Rewrite j(x) = (x+1)(x-1)(x+2)(x-2) from Activity into standard form. Identify the y-intercept before looking back at your graph.
7) How does the y-int relate to the equation? .
Answer: h(x) = (x+1)(x)(x-2) = x3 - x2 - 2x y-intercept is (0,0)
Answer: j(x) = (x+1)(x-1)(x+2)(x-2) = x4 - 5x y-intercept is (0,4)
Answer: The y-intercept is the constant of the polynomial.

55. 12.7
8) What happens to the x-intercepts when the graph doesn’t cross the x-axis or just touches it? What is a possible equation for the graph?
k(x) = (x+1)2(x-2)2
x-intercepts that we study in Algebra 1 are all within the Real Number System. So, the following graph is of a degree 4
equation but only has 2 real-number x-intercepts. In Algebra 2 you will learn about Imaginary Numbers. The 2 “missing” x-intercepts in these cases are imaginary intercepts. We will not study those in this course.
k(x) = (x+1)2 (x-2)2

56. 9) Can you write a function for any squiggle
9) Can you write a function for any squiggle? What do we know about this curve? Can we use the Factor Theorem on this curve?
It crosses the x-axis 8 times in the area that we can see. We cannot be certain it does not cross further on outside our viewing area.
It crosses the y-axis at (0,1). So, It should have a “+1” at the end of the function, if one exists.
It qualifies as a function because each x is used only once.
We cannot be certain that is a polynomial function. So, we cannot automatically apply the Factor Theorem. It’s easier to describe the graph of an equation than the equation of a graph.

57. 12.8
12-8
Objectives : Apply factoring to operations with, and simplification of, rational expressions.

58. Rational Expressions:
The written quotient of two polynomials.
Examples:

59. Lowest Terms
When there is no polynomial that is a factor of its numerator and denominator.
Note: First algebraic expression and it’s lowest term are not equivalent unless x and y not equal to zero.

60. 1) Put the rational expression into lowest terms.
12.8
Factor numerator and denominator
Simplify
1a) Are there any restrictions on n?
Note: Restrictions need to be on original problem.
n ≠ ±½

61. 2) Add
12.8
Factor
Get like terms
Simplify
s ≠ 1 or -3

62. 3) What is the definition of a polynomial? Of a monomial?
12.8
3) What is the definition of a polynomial? Of a monomial?
Polynomial – a monomial or the sum of monomials.
Monomial – a single term in which the exponent for every variable is a positive integer.
4) Is it a rational expression?
4x b) -¾x c)
yes no no

63. 5)Simplify
12.8
Factor
Factor completely
Simplify
r ≠ 0, -5 or -4

64. 6) Add. 12.8 Factor numerator and denominator Simplify each fraction
Get common denominators
Simplify numerators
Combine like terms
n ≠ 1 or -1
Check the graph of the numerator – it does not have x-intercepts 1 or -1

65. Rev 12
Chapter 12 Review
1) Write the (x-4)(x+5)(2x+5) polynomial in standard form.
2)What are the x-ints of the polynomial in Q1?
Answer:
Answer:

66. 3) Simplify. a. -3x(4x2 + 2x -5) b. (y - 4)(y + 2) c. (3r + 1)(3r - 1)
Rev 12
-12x3 - 6x2 + 15x
y2 - 2y - 8
9r difference of 2 squares

67. d. (4x2 - 3x + 2) + (2x2 - 2) e. (5k - 2j) - (2k - 3j + 5)
Rev 12
d. (4x2 - 3x + 2) + (2x2 - 2)
e. (5k - 2j) - (2k - 3j + 5)
f. (4t - 2)(t4 + 3t2 + 4)
6x2 – 3x simplify
3k + j – 5 simplify
4t5 - 2t4 + 12t3 - 6t2 + 16t - 8

68. 5) Know the Perfect Square Patterns and Difference of Two Squares Pattern.
Rev 12
1) Perfect Square Patterns:
(a+b)2 = (a+b)(a+b) = a2 + 2ab + b2
(a-b)2 = (a-b)(a-b) = a2 - 2ab + b2
2) Difference of Two Squares Pattern:
(a+b)(a-b) = a2 - b2
Expand:
(7 + x)2
b) (2n – 6)2
c) (4z + 3a)(4z – 3a)
x2 + 14x + 49
4n2 – 24n + 36
16z2 – 9a2

69. 6) State the multiplication shown in a picture. a.
Rev 12
6) State the multiplication shown in a picture. a. b.
(x+2) (x+1) = x2 + 3x + 2
x · x (x+y) = x3 + x2y

70. 7) Investment situation.
Each birthday from age 15 on, Joan has received $75 from her grandfather. She puts the money into a savings account with a yearly scale factor of p and does not make any withdrawals or additional deposits.
a. Write an expression for the amount Joan will have in the account on her 18th birthday.
b. If the bank pays 6% interest per year, how much will Joan have on her 18th birthday.
Rev 12
75p3 + 75p2 + 75p + 75
75(1.06)3 + 75(1.06)2 + 75(1.06) + 75 = $328.10