factoring module quiz: b
multiple choice: choose the answer that best answers the question, and write the letter in front of the question number.
1.) factor $x^2 + 24x + 80$
a. $(x + 24)(x + 80)$ c. $(x + 5)(x + 16)$
b. $(x + 1)(x + 80)$ d. $(x + 4)(x + 20)$
2.) factor $3x^2 + 2x - 8$
a. $(x - 2)(3x + 4)$ c. $(x - 2)(3x - 4)$
b. $(x + 2)(3x - 4)$ d. $(x + 2)(3x + 4)$
3.) factor $100x^2 - 9z^6$
a. $(10x^2 + 3z^6)(10x^2 - 3z^6)$ c. $(10x + 3z^3)(10x - 3z^3)$
b. $(10x + 3z)(10x - 3z)$ d. already factored
short answer
4.) factor out the gcf $18y^4 + 27y^3 - 6y^2$

Question

factoring module quiz: b
multiple choice: choose the answer that best answers the question, and write the letter in front of the question number.
1.) factor $x^2 + 24x + 80$
	a. $(x + 24)(x + 80)$		c. $(x + 5)(x + 16)$
	b. $(x + 1)(x + 80)$		d. $(x + 4)(x + 20)$
2.) factor $3x^2 + 2x - 8$
	a. $(x - 2)(3x + 4)$		c. $(x - 2)(3x - 4)$
	b. $(x + 2)(3x - 4)$		d. $(x + 2)(3x + 4)$
3.) factor $100x^2 - 9z^6$
	a. $(10x^2 + 3z^6)(10x^2 - 3z^6)$		c. $(10x + 3z^3)(10x - 3z^3)$
	b. $(10x + 3z)(10x - 3z)$		d. already factored
short answer
4.) factor out the gcf $18y^4 + 27y^3 - 6y^2$

Accepted Answer

### Question 1:
# Explanation:
## Step1: Find two numbers that multiply to 80 and add to 24.
We need two numbers $ m $ and $ n $ such that $ m \times n = 80 $ and $ m + n = 24 $.
The factors of 80 are: 1 & 80, 2 & 40, 4 & 20, 5 & 16, 8 & 10.
Among these, 4 and 20 add up to 24 (since $ 4 + 20 = 24 $) and multiply to 80 (since $ 4 \times 20 = 80 $).
## Step2: Factor the quadratic.
Using the numbers 4 and 20, we can factor $ x^2 + 24x + 80 $ as $ (x + 4)(x + 20) $.
# Answer:
d. $(x + 4)(x + 20)$

### Question 2:
# Explanation:
## Step1: Use the AC method for factoring $ ax^2 + bx + c $.
For $ 3x^2 + 2x - 8 $, $ a = 3 $, $ b = 2 $, $ c = -8 $.
First, find two numbers that multiply to $ a \times c = 3 \times (-8) = -24 $ and add to $ b = 2 $.
The numbers are 6 and -4 (since $ 6 \times (-4) = -24 $ and $ 6 + (-4) = 2 $).
## Step2: Rewrite the middle term and factor by grouping.
Rewrite $ 2x $ as $ 6x - 4x $:
$ 3x^2 + 6x - 4x - 8 $
Group the first two and last two terms:
$ (3x^2 + 6x) + (-4x - 8) $
Factor out the GCF from each group:
$ 3x(x + 2) - 4(x + 2) $
Now, factor out $ (x + 2) $:
$ (x + 2)(3x - 4) $
# Answer:
b. $(x + 2)(3x - 4)$

### Question 3:
# Explanation:
## Step1: Recognize the difference of squares.
The expression $ 100x^2 - 9z^6 $ can be written as $ (10x)^2 - (3z^3)^2 $ (since $ 100x^2 = (10x)^2 $ and $ 9z^6 = (3z^3)^2 $).
## Step2: Apply the difference of squares formula $ a^2 - b^2 = (a + b)(a - b) $.
Using $ a = 10x $ and $ b = 3z^3 $, we get:
$ (10x + 3z^3)(10x - 3z^3) $
# Answer:
c. $(10x + 3z^3)(10x - 3z^3)$

### Question 4:
# Explanation:
## Step1: Find the GCF of the coefficients and the variables.
For the coefficients 18, 27, and -6:
The GCF of 18, 27, and 6 is 3 (since $ 18 = 3 \times 6 $, $ 27 = 3 \times 9 $, $ 6 = 3 \times 2 $).
For the variable terms $ y^4 $, $ y^3 $, and $ y^2 $:
The GCF is $ y^2 $ (the lowest power of $ y $).
So, the GCF of $ 18y^4 + 27y^3 - 6y^2 $ is $ 3y^2 $.
## Step2: Factor out the GCF.
Divide each term by $ 3y^2 $:
$ 18y^4 \div 3y^2 = 6y^2 $
$ 27y^3 \div 3y^2 = 9y $
$ -6y^2 \div 3y^2 = -2 $
So, $ 18y^4 + 27y^3 - 6y^2 = 3y^2(6y^2 + 9y - 2) $
# Answer:
$ 3y^2(6y^2 + 9y - 2) $

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QUESTION IMAGE

Question

Explanation:

Question 1:

Step1: Find two numbers that multiply to 80 and add to 24.

Step2: Factor the quadratic.

Step1: Use the AC method for factoring \( ax^2 + bx + c \).

Step2: Rewrite the middle term and factor by grouping.

Step1: Recognize the difference of squares.

Step2: Apply the difference of squares formula \( a^2 - b^2 = (a + b)(a - b) \).

Answer:

Question 2: