Download 4 using a diamond and a generic rectangle

Solutions
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Algebra: 8.1.5 Factoring Shortcuts
Bell Work 1/12 Factor the following polynomials with a diamond and generic rectangle.
a. 7x2 − 20x − 3
−21𝑥 2
−𝟐𝟐𝟐
−20𝑥
𝟏𝟏
−3 −21𝑥 −3
= (𝒙 − 𝟑)(𝟕𝒙 + 𝟏)
A.
𝑥 7𝑥 2
b. 4x2 + 20x + 25
𝟏𝟏𝒙
𝑥
7𝑥
100𝑥 2
1
20𝑥
𝟏𝟎𝒙
= (𝟐𝒙 + 𝟓)𝟐
5 10𝑥
25
2𝑥 4𝑥 2 10𝑥
2𝑥
5
Factor x2 − 6x + 9 using a diamond and a generic rectangle.
= ( 𝒙 − 𝟑) 𝟐
9𝑥 2
−𝟑𝟑
−6𝑥
−3 −3𝑥
−𝟑𝟑
9
𝑥 𝑥 2 −3𝑥
𝑥
−3
This polynomial is a perfect square trinomial. Explain why we call it that.
Because the two factors are equal; its model is shaped like a square; it is binomial squared.
B.
Factor 25x2 – 4 using a diamond and a generic rectangle.
= (𝟓𝟓 + 𝟐)(𝟓𝟓 − 𝟐)
−2 −10𝑥 −4
−100𝑥 2
−𝟏𝟏𝒙
0𝑥
𝟏𝟏𝒙
5𝑥 25𝑥 2 10𝑥
5𝑥
2
What shortcut could you use to factor the difference of squares a2 – b2?
= (𝒂 + 𝒃)(𝒂 − 𝒃)
8-45. Classify each of the following polynomials (completely) and then factor (if possible):
2
Binomial a. x − 49 = (𝒙 + 𝟕)(𝒙 − 𝟕)
Difference of Squares
Trinomial b. x + 2x − 24 = (𝒙 + 𝟔)(𝒙 − 𝟒)
2
Trinomial c. x − 10x + 25 = (𝒙 − 𝟓)
Perfect Square
2
𝟐
Trinomial d. 9x2 + 12x + 4 = (𝟑𝟑 + 𝟐)𝟐
Perfect Square
2
Binomial e. x + 4
This polynomial is Prime
Binomial f. 4x2 − 25 = (𝟐𝟐 + 𝟓)(𝟐𝟐 − 𝟓)
Difference of Squares
6 6𝑥 −24
𝑥 𝑥 2 −4𝑥
𝑥
2 6𝑥
3𝑥 9𝑥 2
3𝑥
−4
4
6𝑥
2
8-46. Classify each of the following polynomials (completely) and then factor (if possible).
2
Binomial a. 25x − 1 = (𝟓𝟓 + 𝟏)(𝟓𝟓 − 𝟏)
Difference of Squares
Trinomial b. x2 − 5x − 36 = (𝒙 − 𝟗)(𝒙 + 𝟒)
Trinomial c. x2 + 8x + 16 = (𝒙 + 𝟒)𝟐
Perfect Square
Trinomial d. 9x2 − 12x + 4 = (𝟑𝟑 − 𝟐)𝟐
Perfect Square
2
Binomial e. 9x + 4
+4 4𝑥 −36
𝑥 𝑥 2 −9𝑥
𝑥
4 4𝑥
−9
𝑥 𝑥2
𝑥
16
4𝑥
4
−2 −6𝑥
4
3𝑥 9𝑥 2 −6𝑥
3𝑥
This polynomial is Prime
Binomial f. 9x2 − 100 = (𝟑𝒙 + 𝟏𝟏)(𝟑𝟑 − 𝟏𝟏)
Difference of Squares
−2
8-49. Factor each polynomial completely.
a. x2 − 64 = (𝒙 + 𝟖)(𝒙 − 𝟖)
b. y2 − 6y + 9 = (𝒚 − 𝟑)𝟐
2
c. 4x + 4x + 1 = (𝟐𝟐 + 𝟏)
𝟐
−3 −3𝑦
9
𝑦 𝑦 2 −3𝑦
𝑦
−3
d. 5x − 45 = 𝟓(𝒙𝟐 − 𝟗) = 𝟓(𝒙 + 𝟑)(𝒙 − 𝟑)
2
1 2𝑥
2𝑥 4𝑥 2
2𝑥
1
2𝑥
1
8-50. Simplify each expression below. Your answer should contain no parentheses and no
negative exponents.
2
1
a. �− 𝑥 5 𝑦 3 �
3
0
=1
Anything to the zero power (except for 0) is 1.
1
b. �252 𝑥 5 � (4𝑥 −6 ) = (5𝑥 5 ) (4𝑥 −6 ) = 20𝑥 −1 =
c. 5𝑡 −3 =
d.
𝟓
𝒕𝟑
1
= (𝑥 6 𝑦 3 )3 = 𝒙𝟐 𝒚
𝟐𝟐
𝒙
8-51. Solve each system algebraically. Confirm your solutions by graphing.
a.
y = 4x + 5
y = −2x − 13
4𝑥 + 5 = −2𝑥 − 13
𝑦 = 4(−3) + 5
6𝑥 = −18
𝑥 = −3
𝑦 = −7
𝑦 = −12 + 5
6𝑥 + 5 = −13
b. 2x + y = 9
y = −x + 4
(−𝟑, −𝟕)
𝑦 = −(5) + 4
2𝑥 + −𝑥 + 4 = 9
𝑥+4= 9
(𝟓, −𝟏)
𝑦 = −1
𝑥=5
8-52. Consider the sequence 4, 8, …
If the sequence is arithmetic, write the
first 4 terms and an equation
4, 8, 12, 16, …
If the sequence is geometric, write the
first 4 terms and an equation
4, 8, 16, 32, …
a.
b.
c.
Create another sequence that is
neither arithmetic nor geometric and
still starts with 4, 8, ….
𝑡 (𝑛) = 4𝑛
𝑡(𝑛) = 2(2)𝑛
4, 8, 13,19, …
8-53. Solve the following equations for x.
a.
b.
c.
d.
4x − 6y = 20
1
2
4
5
(x − 6) = 9
+
18
𝑥
=8
4𝑥 = 20 + 6𝑦
𝟑
𝒙 = 𝟓+ 𝒚
𝟐
2 + |2𝑥 − 3| = 5
𝑥 − 6 = 18
|2𝑥 − 3| = 3
𝒙 = 𝟐𝟐
2𝑥 − 3 = 3 𝑜𝑜 2𝑥 − 3 = −3
2𝑥 = 6 𝑜𝑜 2𝑥 = 0
𝒙 = 𝟑 𝒐𝒐 𝒙 = 𝟎
4 18
5𝑥 � +
= 8�
5 𝑥
4𝑥 + 90 = 40𝑥
90 = 36𝑥
2.5 𝑜𝑜
5
=𝑥
2
C. Factor and graph each quadratic equation. What are the root(s), vertex and y-intercept(s) for each?
i.
5 5𝑥
𝑥 𝑥2
𝑥
y = x2 + 6x + 5
ii.
y = -x2 + 9
5
1𝑥
1
𝑦 = (𝑥 + 5)(𝑥 + 1)
Roots: -5 and -1
Vertex = (-3, 4)
𝑦 – intercept = 5
EOC Practice:
𝒚 = −(𝒙𝟐 − 𝟗)
𝑦 = −(𝑥 + 3)(𝑥 − 3)
Roots: -3 and 3
Vertex = (0, 9)
𝑦 – intercept = 9