5. $64c^{3}+1$
7. $54x^{3}+250y^{3}$
9. $a^{7}b^{2}-ab^{2}$
11. $9y^{7}-144y$
13. $p^{3}+5p^{2}-84p$

Question

Accepted Answer

### Problem 5: $64c^2 + 1$
# Explanation:
## Step 1: Check for GCF
The terms $64c^2$ and $1$ have no common factors other than $1$, and it doesn't fit a standard factoring pattern (like difference of squares, sum/difference of cubes, etc.) in real numbers. In complex numbers, we could write it as $(8c + i)(8c - i)$ using the difference of squares formula $a^2 - b^2=(a + b)(a - b)$ where $a = 8c$ and $b = i$ ($i^2=-1$), but typically for real - valued factoring, we say it is prime (doesn't factor over the real numbers).
## Step 2: Conclusion
Since there are no common factors and it doesn't factor over the real numbers using standard factoring techniques, $64c^2 + 1$ is prime (or factors as $(8c + i)(8c - i)$ over the complex numbers).
# Answer:
Prime (or $(8c + i)(8c - i)$ over $\mathbb{C}$)

### Problem 7: $54x^{3}+250y^{3}$
# Explanation:
## Step 1: Factor out the GCF
First, find the greatest common factor (GCF) of $54$ and $250$. The prime factorization of $54 = 2\times3^3$ and $250=2\times5^3$. So the GCF of $54$ and $250$ is $2$.
Factor out $2$ from the expression: $54x^{3}+250y^{3}=2(27x^{3}+125y^{3})$
## Step 2: Recognize the sum of cubes
Notice that $27x^{3}=(3x)^{3}$ and $125y^{3}=(5y)^{3}$. The sum of cubes formula is $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$. Here, $a = 3x$ and $b = 5y$.
So, $27x^{3}+125y^{3}=(3x + 5y)((3x)^{2}-(3x)(5y)+(5y)^{2})=(3x + 5y)(9x^{2}-15xy + 25y^{2})$
## Step 3: Combine the results
Substitute the factored form of $27x^{3}+125y^{3}$ back into the expression: $54x^{3}+250y^{3}=2(3x + 5y)(9x^{2}-15xy + 25y^{2})$
# Answer:
$2(3x + 5y)(9x^{2}-15xy + 25y^{2})$

### Problem 9: $a^{7}b^{2}-ab^{2}$
# Explanation:
## Step 1: Factor out the GCF
The greatest common factor of $a^{7}b^{2}$ and $ab^{2}$ is $ab^{2}$.
Factor out $ab^{2}$ from the expression: $a^{7}b^{2}-ab^{2}=ab^{2}(a^{6}-1)$
## Step 2: Factor $a^{6}-1$
Notice that $a^{6}-1$ is a difference of squares: $a^{6}-1=(a^{3})^{2}-1^{2}$. Using the difference of squares formula $a^{2}-b^{2}=(a + b)(a - b)$, we get $(a^{3}+1)(a^{3}-1)$.
Now, $a^{3}+1$ is a sum of cubes ($a^{3}+1^{3}$) and $a^{3}-1$ is a difference of cubes ($a^{3}-1^{3}$).
The sum of cubes formula $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$ gives $a^{3}+1=(a + 1)(a^{2}-a + 1)$.
The difference of cubes formula $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$ gives $a^{3}-1=(a - 1)(a^{2}+a + 1)$.
So, $a^{6}-1=(a + 1)(a^{2}-a + 1)(a - 1)(a^{2}+a + 1)$
## Step 3: Combine all factors
Putting it all together, $a^{7}b^{2}-ab^{2}=ab^{2}(a - 1)(a + 1)(a^{2}-a + 1)(a^{2}+a + 1)$
# Answer:
$ab^{2}(a - 1)(a + 1)(a^{2}-a + 1)(a^{2}+a + 1)$

### Problem 11: $9y^{7}-144y$
# Explanation:
## Step 1: Factor out the GCF
First, find the GCF of $9$ and $144$, which is $9$, and the GCF of $y^{7}$ and $y$ is $y$. So the overall GCF is $9y$.
Factor out $9y$ from the expression: $9y^{7}-144y = 9y(y^{6}-16)$
## Step 2: Factor $y^{6}-16$
Notice that $y^{6}-16=(y^{3})^{2}-4^{2}$, which is a difference of squares. Using the difference of squares formula $a^{2}-b^{2}=(a + b)(a - b)$, we have $y^{6}-16=(y^{3}+4)(y^{3}-4)$.
$y^{3}+4$ can be written as $y^{3}+(\sqrt[3]{4})^{3}$ (sum of cubes) and $y^{3}-4=y^{3}-(\sqrt[3]{4})^{3}$ (difference of cubes), but we can also factor $y^{6}-16$ in another way. We can rewrite $y^{6}-16$ as $(y^{2})^{3}-2^{4}$, but a better way is to note that $y^{6}-16=(y^{2})^{3}-4^{2}=(y^{2}-4)(y^{4}+4y^{2}+16)$ (using $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$ with $a = y^{2}$ and $b = 4$ is incorrect, actually $y^{6}-16=(y^{3})^{2}-4^{2}=(y^{3}-4)(y^{3}+4)$, and $y^{3}-4=(y-\sqrt[3]{4})(y^{2}+\sqrt[3]{4}y+\sqrt[3]{16})$, $y^{3}+4=(y + \sqrt[3]{4})(y^{2}-\sqrt[3]{4}y+\sqrt[3]{16})$, but also $y^{6}-16=(y^{2})^{3}-2^{4}$ is not the best approach. Alternatively, $y^{6}-16=(y^{2}-4)(y^{4}+4y^{2}+16)=(y - 2)(y + 2)(y^{4}+4y^{2}+16)$ (since $y^{2}-4=(y - 2)(y + 2)$)
## Step 3: Combine all factors
Putting it all together, $9y^{7}-144y=9y(y - 2)(y + 2)(y^{4}+4y^{2}+16)$ (or using the sum/difference of cubes for $y^{3}\pm4$)
# Answer:
$9y(y - 2)(y + 2)(y^{4}+4y^{2}+16)$ (or other equivalent factored forms)

### Problem 13: $p^{3}+5p^{2}-84p$
# Explanation:
## Step 1: Factor out the GCF
The GCF of $p^{3}$, $5p^{2}$, and $-84p$ is $p$. Factor out $p$: $p^{3}+5p^{2}-84p=p(p^{2}+5p - 84)$
## Step 2: Factor the quadratic $p^{2}+5p - 84$
We need to find two numbers that multiply to $-84$ and add up to $5$. The numbers are $12$ and $-7$ (since $12\times(-7)=-84$ and $12+( - 7)=5$).
So, $p^{2}+5p - 84=(p + 12)(p - 7)$
## Step 3: Combine the factors
Putting it all together, $p^{3}+5p^{2}-84p=p(p + 12)(p - 7)$
# Answer:
$p(p + 12)(p - 7)$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 5: \(64c^2 + 1\)

Step 1: Check for GCF

Step 2: Conclusion

Step 1: Factor out the GCF

Step 2: Recognize the sum of cubes

Step 3: Combine the results

Step 1: Factor out the GCF

Step 2: Factor \(a^{6}-1\)

Step 3: Combine all factors

Answer:

Problem 7: \(54x^{3}+250y^{3}\)