date: ______________________ per: ______ homework 2: factoring polynomials
directions: complete the following rules.
1. difference of squares
$a^2 - b^2 =$
2. sum of cubes
$a^3 + b^3 =$
3. difference of cubes
$a^3 - b^3 =$
directions: factor each polynomial completely. make sure to check for a gcf first.
4. $x^4 - 36$
5. $64c^3 + 1$
6. $k^3 - 27$
7. $54x^3 + 250y^3$
8. $3m^4 - 48n^2$
9. $a^7b^2 - ab^2$
10. $x^3y^2 - 343y^5$
11. $9y^7 - 144y$
12. $w^2 - 13w + 36$
13. $p^3 + 5p^2 - 84p$

Question

Accepted Answer

### Problem 1: Difference of Squares
# Explanation:
## Step1: Recall the formula for difference of squares
The formula for the difference of squares is $a^2 - b^2=(a + b)(a - b)$
# Answer:
$(a + b)(a - b)$

### Problem 2: Sum of Cubes
# Explanation:
## Step1: Recall the formula for sum of cubes
The formula for the sum of cubes is $a^3 + b^3=(a + b)(a^2 - ab + b^2)$
# Answer:
$(a + b)(a^2 - ab + b^2)$

### Problem 3: Difference of Cubes
# Explanation:
## Step1: Recall the formula for difference of cubes
The formula for the difference of cubes is $a^3 - b^3=(a - b)(a^2 + ab + b^2)$
# Answer:
$(a - b)(a^2 + ab + b^2)$

### Problem 4: Factor $x^4-36$
# Explanation:
## Step1: Recognize as difference of squares
We can write $x^4-36=(x^2)^2 - 6^2$
## Step2: Apply difference of squares formula
Using $a^2 - b^2=(a + b)(a - b)$ where $a = x^2$ and $b = 6$, we get $(x^2 + 6)(x^2 - 6)$
## Step3: Check if further factoring is possible
$x^2+6$ cannot be factored over real numbers, but $x^2 - 6=(x+\sqrt{6})(x - \sqrt{6})$ if we consider real numbers. But if we are factoring over integers, we stop at $(x^2 + 6)(x^2 - 6)$
# Answer:
$(x^2 + 6)(x^2 - 6)$ (or $(x^2 + 6)(x+\sqrt{6})(x - \sqrt{6})$ for real numbers)

### Problem 5: Factor $64c^3 + 1$
# Explanation:
## Step1: Recognize as sum of cubes
We can write $64c^3+1=(4c)^3+1^3$
## Step2: Apply sum of cubes formula
Using $a^3 + b^3=(a + b)(a^2 - ab + b^2)$ where $a = 4c$ and $b = 1$, we get $(4c + 1)((4c)^2-4c\times1 + 1^2)=(4c + 1)(16c^2-4c + 1)$
# Answer:
$(4c + 1)(16c^2-4c + 1)$

### Problem 6: Factor $k^3-27$
# Explanation:
## Step1: Recognize as difference of cubes
We can write $k^3 - 27=k^3-3^3$
## Step2: Apply difference of cubes formula
Using $a^3 - b^3=(a - b)(a^2 + ab + b^2)$ where $a = k$ and $b = 3$, we get $(k - 3)(k^2+3k + 9)$
# Answer:
$(k - 3)(k^2+3k + 9)$

### Problem 7: Factor $54x^3 + 250y^3$
# Explanation:
## Step1: Factor out the GCF
First, we find the GCF of 54 and 250. The GCF of 54 and 250 is 2. So we factor out 2: $54x^3+250y^3 = 2(27x^3+125y^3)$
## Step2: Recognize as sum of cubes
We can write $27x^3+125y^3=(3x)^3+(5y)^3$
## Step3: Apply sum of cubes formula
Using $a^3 + b^3=(a + b)(a^2 - ab + b^2)$ where $a = 3x$ and $b = 5y$, we get $2[(3x + 5y)((3x)^2-3x\times5y+(5y)^2)]=2(3x + 5y)(9x^2-15xy + 25y^2)$
# Answer:
$2(3x + 5y)(9x^2-15xy + 25y^2)$

### Problem 8: Factor $3m^4-48n^2$
# Explanation:
## Step1: Factor out the GCF
First, we find the GCF of 3 and 48, which is 3. So we factor out 3: $3m^4-48n^2=3(m^4 - 16n^2)$
## Step2: Recognize as difference of squares
We can write $m^4-16n^2=(m^2)^2-(4n)^2$
## Step3: Apply difference of squares formula
Using $a^2 - b^2=(a + b)(a - b)$ where $a = m^2$ and $b = 4n$, we get $3(m^2 + 4n)(m^2 - 4n)$
## Step4: Check for further factoring
$m^2 + 4n$ cannot be factored over integers, but $m^2 - 4n=(m + 2\sqrt{n})(m - 2\sqrt{n})$ if we consider real numbers (assuming $n\geq0$). Over integers, we stop at $3(m^2 + 4n)(m^2 - 4n)$
# Answer:
$3(m^2 + 4n)(m^2 - 4n)$ (or $3(m^2 + 4n)(m + 2\sqrt{n})(m - 2\sqrt{n})$ for real numbers)

### Problem 9: Factor $a^7b^2 - ab^2$
# Explanation:
## Step1: Factor out the GCF
First, we factor out the GCF which is $ab^2$: $a^7b^2 - ab^2=ab^2(a^6 - 1)$
## Step2: Recognize $a^6 - 1$ as difference of squares
We can write $a^6 - 1=(a^3)^2-1^2$
## Step3: Apply difference of squares formula
Using $a^2 - b^2=(a + b)(a - b)$ where $a = a^3$ and $b = 1$, we get $ab^2(a^3 + 1)(a^3 - 1)$
## Step4: Factor $a^3 + 1$ and $a^3 - 1$ using sum and difference of cubes
$a^3 + 1=(a + 1)(a^2 - a + 1)$ and $a^3 - 1=(a - 1)(a^2 + a + 1)$
So the fully factored form is $ab^2(a + 1)(a^2 - a + 1)(a - 1)(a^2 + a + 1)$
# Answer:
$ab^2(a + 1)(a^2 - a + 1)(a - 1)(a^2 + a + 1)$

### Problem 10: Factor $x^3y^2-343y^5$
# Explanation:
## Step1: Factor out the GCF
First, we factor out the GCF which is $y^2$: $x^3y^2-343y^5=y^2(x^3 - 343y^3)$
## Step2: Recognize as difference of cubes
We can write $x^3 - 343y^3=x^3-(7y)^3$
## Step3: Apply difference of cubes formula
Using $a^3 - b^3=(a - b)(a^2 + ab + b^2)$ where $a = x$ and $b = 7y$, we get $y^2(x - 7y)(x^2+7xy + 49y^2)$
# Answer:
$y^2(x - 7y)(x^2+7xy + 49y^2)$

### Problem 11: Factor $9y^7-144y$
# Explanation:
## Step1: Factor out the GCF
First, we find the GCF of 9 and 144, which is 9, and also factor out $y$. So we factor out $9y$: $9y^7-144y=9y(y^6 - 16)$
## Step2: Recognize $y^6 - 16$ as difference of squares
We can write $y^6 - 16=(y^3)^2-4^2$
## Step3: Apply difference of squares formula
Using $a^2 - b^2=(a + b)(a - b)$ where $a = y^3$ and $b = 4$, we get $9y(y^3 + 4)(y^3 - 4)$
## Step4: Check for further factoring
$y^3 + 4$ and $y^3 - 4$ can be factored using sum and difference of cubes (if we consider real numbers). $y^3 + 4=(y+\sqrt[3]{4})(y^2-\sqrt[3]{4}y+\sqrt[3]{16})$ and $y^3 - 4=(y-\sqrt[3]{4})(y^2+\sqrt[3]{4}y+\sqrt[3]{16})$
So the fully factored form (over real numbers) is $9y(y+\sqrt[3]{4})(y^2-\sqrt[3]{4}y+\sqrt[3]{16})(y-\sqrt[3]{4})(y^2+\sqrt[3]{4}y+\sqrt[3]{16})$
# Answer:
$9y(y^3 + 4)(y^3 - 4)$ (or $9y(y+\sqrt[3]{4})(y^2-\sqrt[3]{4}y+\sqrt[3]{16})(y-\sqrt[3]{4})(y^2+\sqrt[3]{4}y+\sqrt[3]{16})$ for real numbers)

### Problem 12: Factor $w^2-13w + 36$
# Explanation:
## Step1: Find two numbers that multiply to 36 and add to - 13
We need two numbers $m$ and $n$ such that $m\times n=36$ and $m + n=-13$. The numbers are - 4 and - 9 since $(-4)\times(-9)=36$ and $(-4)+(-9)=-13$
## Step2: Factor the quadratic
Using the formula $x^2+(m + n)x+mn=(x + m)(x + n)$, we get $(w - 4)(w - 9)$
# Answer:
$(w - 4)(w - 9)$

### Problem 13: Factor $p^3+5p^2-84p$
# Explanation:
## Step1: Factor out the GCF
First, we factor out the GCF which is $p$: $p^3+5p^2-84p=p(p^2 + 5p - 84)$
## Step2: Factor the quadratic $p^2 + 5p - 84$
We need two numbers that multiply to - 84 and add to 5. The numbers are 12 and - 7 since $12\times(-7)=-84$ and $12+(-7)=5$
## Step3: Factor the quadratic
Using the formula $x^2+(m + n)x+mn=(x + m)(x + n)$, we get $p(p + 12)(p - 7)$
# Answer:
$p(p + 12)(p - 7)$

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QUESTION IMAGE

Question

Explanation:

Problem 1: Difference of Squares

Step1: Recall the formula for difference of squares

Step1: Recall the formula for sum of cubes

Step1: Recall the formula for difference of cubes

Answer:

Problem 2: Sum of Cubes