name emily tulley teacher hour 2ed unit 5 factoring and polynomial equations unit 5 test: a 1. choose true or false for each statement about factoring $4x^{2}y - 16y$. a.) you can factor using a perfect square trinomial pattern. ____true. ____false b.) you can factor by grouping. ____true. ____false c.) you can factor out the greatest monomial factor. ____true. ____false d.) you can factor using the difference of two squares pattern. ____true. ____false 2. using any factoring method, can you factor each of the following polynomials? a.) $8x^{3}+9$ ____yes ____no b.) $x^{3}+10$ ____yes ____no c.) $x^{2}-8x - 12$ ____yes ____no d.) $x^{2}-10x - 11$ ____yes ____no e.) $x^{6}-4$ ____yes ____no for 3–4, factor the polynomial completely, or identify it as irreducible. 3. $x^{2}+4x - 21$ 4. $16x^{2}+40x + 25$ ginal content copyright © by houghton mifflin harcourt. additions and changes to the original content are the responsibility of the

Question

Accepted Answer

### Question 1
#### Part A
# Explanation:
## Step1: Analyze the polynomial $4x^{2}y - 16y$
The polynomial has two terms, a perfect square trinomial pattern requires three terms (of the form $a^{2}+2ab + b^{2}$ or $a^{2}-2ab + b^{2}$). So we can't factor using a perfect square trinomial pattern.
# Answer:
False

#### Part B
# Explanation:
## Step1: Analyze factoring by grouping
Factoring by grouping is used for polynomials with four terms (grouping two terms at a time). The given polynomial $4x^{2}y-16y$ has two terms, so we can't factor by grouping.
# Answer:
False

#### Part C
# Explanation:
## Step1: Find the greatest monomial factor
The greatest monomial factor of $4x^{2}y$ and $- 16y$ is $4y$. We can factor out $4y$ from the polynomial: $4x^{2}y-16y=4y(x^{2} - 4)$
# Answer:
True

#### Part D
# Explanation:
## Step1: Factor the polynomial after factoring out GCF
After factoring out $4y$ from $4x^{2}y - 16y$, we get $4y(x^{2}-4)$. And $x^{2}-4$ is a difference of two squares ($x^{2}-4=x^{2}-2^{2}$) and can be factored as $(x + 2)(x - 2)$ using the difference of two squares pattern $a^{2}-b^{2}=(a + b)(a - b)$
# Answer:
True

### Question 2
#### Part A
# Explanation:
## Step1: Analyze the polynomial $8x^{3}+9$
The polynomial $8x^{3}+9=(2x)^{3}+3^{3}$. The sum of cubes formula is $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$. Here $a = 2x$ and $b=3$, so $8x^{3}+9=(2x + 3)(4x^{2}-6x + 9)$
# Answer:
Yes

#### Part B
# Explanation:
## Step1: Analyze the polynomial $x^{3}+10$
The polynomial $x^{3}+10=x^{3}+(\sqrt[3]{10})^{3}$. Using the sum of cubes formula $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$ where $a=x$ and $b = \sqrt[3]{10}$, so $x^{3}+10=(x+\sqrt[3]{10})(x^{2}-\sqrt[3]{10}x+\sqrt[3]{100})$
# Answer:
Yes (If we consider factoring over real numbers. If we consider factoring over integers, since $10$ is not a perfect cube of an integer, but in general factoring (over real numbers) is possible)

#### Part C
# Explanation:
## Step1: Analyze the quadratic $x^{2}-8x - 12$
For a quadratic $ax^{2}+bx + c$ ($a = 1$, $b=-8$, $c=-12$), we check the discriminant $D=b^{2}-4ac=(-8)^{2}-4\times1\times(-12)=64 + 48=112$. The roots of the quadratic equation $x^{2}-8x - 12=0$ are $x=\frac{8\pm\sqrt{112}}{2}=\frac{8\pm4\sqrt{7}}{2} = 4\pm2\sqrt{7}$. Since the roots are irrational, but we can write the quadratic as $(x-(4 + 2\sqrt{7}))(x-(4 - 2\sqrt{7}))$. But if we consider factoring over integers, we need two integers $m$ and $n$ such that $m + n=-8$ and $mn=-12$. The pairs of factors of $- 12$ are $(-12,1)$ sum is $-11$, $(-6,2)$ sum is $-4$, $(-4,3)$ sum is $-1$, $(12,-1)$ sum is $11$, $(6,-2)$ sum is $4$, $(4,-3)$ sum is $1$. None of the pairs sum to $-8$. But if we consider factoring over real numbers, it is factorable. However, usually, in the context of basic factoring (over integers), we say no. But if we consider factoring over real numbers, we say yes. But since the problem is likely considering factoring over integers, we check the integer factors. Since we can't find two integers $m$ and $n$ such that $m + n=-8$ and $mn=-12$, we can't factor it over integers. But if we consider factoring over real numbers, we can. But the answer depends on the context. If we assume factoring over integers, the answer is No. If we assume factoring over real numbers, the answer is Yes. But in the context of a typical middle school or high school algebra problem, factoring over integers is assumed. So we will go with No.
# Answer:
No

#### Part D
# Explanation:
## Step1: Analyze the polynomial $x^{2}-10x - 11$
We need two integers $m$ and $n$ such that $m + n=-10$ and $mn=-11$. The factors of $- 11$ are $-11$ and $1$ since $-11+1=-10$ and $(-11)\times1=-11$. So $x^{2}-10x - 11=(x-11)(x + 1)$
# Answer:
Yes

#### Part E
# Explanation:
## Step1: Analyze the polynomial $x^{6}-4$
We can rewrite $x^{6}-4=(x^{3})^{2}-2^{2}$, which is a difference of two squares. Using the difference of two squares formula $a^{2}-b^{2}=(a + b)(a - b)$ with $a=x^{3}$ and $b = 2$, we get $x^{6}-4=(x^{3}+2)(x^{3}-2)$. Also, $x^{3}+2=(x+\sqrt[3]{2})(x^{2}-\sqrt[3]{2}x+\sqrt[3]{4})$ and $x^{3}-2=(x-\sqrt[3]{2})(x^{2}+\sqrt[3]{2}x+\sqrt[3]{4})$ (using sum and difference of cubes formula)
# Answer:
Yes

### Question 3
# Explanation:
## Step1: Analyze the quadratic $x^{2}+4x-21$
We need two integers $m$ and $n$ such that $m + n = 4$ and $mn=-21$. The factors of $-21$ are $7$ and $- 3$ since $7+( - 3)=4$ and $7\times(-3)=-21$
## Step2: Factor the quadratic
So $x^{2}+4x - 21=(x + 7)(x-3)$
# Answer:
$(x + 7)(x - 3)$

### Question 4
# Explanation:
## Step1: Analyze the quadratic $16x^{2}+40x + 25$
The quadratic is of the form $a^{2}+2ab + b^{2}$ where $a = 4x$ ($(4x)^{2}=16x^{2}$) and $b = 5$ ($5^{2}=25$) and $2ab=2\times4x\times5 = 40x$
## Step2: Factor using perfect square trinomial formula
Using the formula $a^{2}+2ab + b^{2}=(a + b)^{2}$, we get $16x^{2}+40x + 25=(4x + 5)^{2}$
# Answer:
$(4x + 5)^{2}$

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QUESTION IMAGE

Question

Explanation:

Question 1

Part A

Step1: Analyze the polynomial \(4x^{2}y - 16y\)

Step1: Analyze factoring by grouping

Step1: Find the greatest monomial factor

Answer:

Part B