Algebra 2 worksheet focusing on factoring polynomials by grouping, including problems like x² + xy + x + y and 2x² - 4x + 6xy - 12y.

Here are the solutions for the factoring problems on the worksheet. The method used here is factoring by grouping, which involves splitting the terms into pairs, finding the greatest common factor (GCF) for each pair, and then factoring out the common binomial.

1) $3(x + y) + q(x + y)$
* Identify the common group: Both parts contain $(x + y)$.
* Factor it out: The remaining parts are $3$ and $q$.
* Result: $(x + y)(3 + q)$

2) $w(n - y) - 8(n - y)$
* Identify the common group: Both parts contain $(n - y)$.
* Factor it out: The remaining parts are $w$ and $-8$.
* Result: $(n - y)(w - 8)$

3) $3ab(x - y) + y(y - x)$
* Notice that $(x - y)$ and $(y - x)$ are opposites. We can rewrite $(y - x)$ as $-(x - y)$.
* Rewrite the expression: $3ab(x - y) - y(x - y)$.
* Factor out $(x - y)$: The remaining parts are $3ab$ and $-y$.
* Result: $(x - y)(3ab - y)$

4) $3p(2q - p) - 2q(p - 2q)$
* Notice that $(2q - p)$ and $(p - 2q)$ are opposites. Rewrite $(p - 2q)$ as $-(2q - p)$.
* Rewrite the expression: $3p(2q - p) + 2q(2q - p)$ (since minus a negative becomes positive).
* Factor out $(2q - p)$: The remaining parts are $3p$ and $2q$.
* Result: $(2q - p)(3p + 2q)$

5) $2a(b + 3) - (3 + b)$
* Recall that addition is commutative, so $(b + 3)$ is the same as $(3 + b)$.
* Treat $-(3 + b)$ as $-1(3 + b)$.
* Factor out $(3 + b)$: The remaining parts are $2a$ and $-1$.
* Result: $(3 + b)(2a - 1)$

6) $3a + ab + 3c + bc$
* Group the first two and last two terms: $(3a + ab) + (3c + bc)$.
* Factor GCF from first group ($a$): $a(3 + b)$.
* Factor GCF from second group ($c$): $c(3 + b)$.
* Factor out the common binomial $(3 + b)$.
* Result: $(3 + b)(a + c)$

7) $x^2 - 2x + xy - 2y$
* Group terms: $(x^2 - 2x) + (xy - 2y)$.
* Factor GCF from first group ($x$): $x(x - 2)$.
* Factor GCF from second group ($y$): $y(x - 2)$.
* Factor out the common binomial $(x - 2)$.
* Result: $(x - 2)(x + y)$

8) $a^2 - 2a + ay - 2y$
* Group terms: $(a^2 - 2a) + (ay - 2y)$.
* Factor GCF from first group ($a$): $a(a - 2)$.
* Factor GCF from second group ($y$): $y(a - 2)$.
* Factor out the common binomial $(a - 2)$.
* Result: $(a - 2)(a + y)$

9) $a^2 - 2ay + 6a - 12y$
* Group terms: $(a^2 - 2ay) + (6a - 12y)$.
* Factor GCF from first group ($a$): $a(a - 2y)$.
* Factor GCF from second group ($6$): $6(a - 2y)$.
* Factor out the common binomial $(a - 2y)$.
* Result: $(a - 2y)(a + 6)$

10) $p^2 - 2pr + 6p - 12r$
* Group terms: $(p^2 - 2pr) + (6p - 12r)$.
* Factor GCF from first group ($p$): $p(p - 2r)$.
* Factor GCF from second group ($6$): $6(p - 2r)$.
* Factor out the common binomial $(p - 2r)$.
* Result: $(p - 2r)(p + 6)$

11) $2a^2 + 6a + a + 2$
* *Note: This problem appears to have a typo in the original text ($2a^2 + 7a + 2$ does not factor cleanly by grouping with integers). Assuming the intended middle terms allow for grouping, let's look at the structure.*
* If we assume the question meant $2a^2 + 4a + a + 2$ (which sums to $2a^2+5a+2$), the grouping would be:
* $2a(a+2) + 1(a+2) \rightarrow (a+2)(2a+1)$.
* However, sticking strictly to the printed text $2a^2 + 6a + a + 2 = 2a^2 + 7a + 2$, this is prime (cannot be factored using simple integer grouping).
* *Alternative Interpretation:* Often in these worksheets, a term like $6a$ might be paired with a constant. Let's try grouping $(2a^2 + a) + (6a + 2)$.
* $a(2a + 1) + 2(3a + 1)$. No common binomial.
* Let's try grouping $(2a^2 + 6a) + (a + 2)$.
* $2a(a + 3) + 1(a + 2)$. No common binomial.
* Correction: Looking closely at typical algebra patterns, Problem 11 is likely $2a^2 + 4a + 3a + 6$ or similar. But based strictly on the image text "$2a^2 + 6a + a + 2$", it doesn't factor nicely.
* *Wait, looking at Problem 12 and 13, they follow a pattern.* Let's re-read #11 carefully. It says $2a^2 + 6a + a + 2$.
* Actually, if we rearrange: $2a^2 + a + 6a + 2$.
* $a(2a + 1) + 2(3a + 1)$. Still no match.
* Let's assume there is a typo in the book and it should have been $2a^2 + 4a + 3a + 6$ -> $(2a+3)(a+2)$.
* OR $2a^2 + 2a + 6a + 6$ -> $2a(a+1)+6(a+1) \rightarrow 2(a+1)(a+3)$.
* Given the ambiguity, I will provide the answer for the most likely intended "clean" version close to the text: $2a^2 + 5a + 2$ factors to $(2a + 1)(a + 2)$. (This assumes the $6a$ was a typo for $4a$ or the $a$ was a typo for $3a$ etc).
* *However*, if we must solve exactly what is written, it is Prime (not factorable over integers).
* *Let's look at #12 for context.*

12) $p^2 - 2pq + qp - 2qr$
* Simplify middle terms: $-2pq + qp = -pq$. So, $p^2 - pq - 2qr$. This doesn't look right for grouping either.
* Let's re-read the image text for #12: $p^2 - 2pq + qp - 2qr$? No, it looks like $p^2 - 2pq + qp - 2q^2$? Or maybe $p^2 - 2pq + 3qp ...$?
* Let's look really closely at crop 5 and 6.
* 11) $2a^2 + 6a + a + 2$
* 12) $p^2 - 2pq + qp - 2qr$ ?? The last term is hard to read. It looks like $2qr$ or $2q^2$.
* Let's assume standard grouping patterns.
* If #12 is $p^2 - 2pq + qp - 2q^2$:
* Combine middle: $-pq$. $p^2 - pq - 2q^2$. Factors to $(p-2q)(p+q)$.
* If #12 is $p^2 - 2pq + 2qp - 4q^2$?
* Let's try grouping as written in similar problems:
* Group 1: $p^2 - 2pq = p(p - 2q)$
* Group 2: Need something that factors to $k(p - 2q)$.
* If the last two terms were $+ ap - 2aq$, it would work.
* The text says $+ qp - 2qr$. This implies variables $p, q, r$.
* $p(p - 2q) + q(p - 2r)$. No common factor.
* Likely Typo in Book: Problem 12 probably meant $p^2 - 2pq + 3p - 6q$ or similar.
* Let's look at Problem 13: $ab - 3ay - 6m + 18my$? No, $ab - 3ay - 6b + 18by$?
* Text: $ab - 3ay - 6m + 18my$? The letters are blurry.
* Let's assume the pattern holds: $A(B-C) + D(B-C)$.

*Let's restart the difficult ones (11-15) by assuming standard textbook typos where the grouping MUST work.*

Re-evaluating 11: $2a^2 + 6a + a + 2$.
If we group $(2a^2 + a) + (6a + 2) \rightarrow a(2a+1) + 2(3a+1)$. Fail.
If we group $(2a^2 + 2) + (6a + a)$. Fail.
*Most likely intended problem:* $2a^2 + 4a + 3a + 6$ $\rightarrow (2a+3)(a+2)$.
*Or:* $2a^2 + 2a + 6a + 6$ $\rightarrow 2(a+1)(a+3)$.
Given the visual "6a", maybe it's $2a^2 + 6a + 5a + 15$? $\rightarrow (2a+5)(a+3)$.
Without clearer text, I will mark 11 as likely containing a typo but provide the factorization for $2a^2+7a+2$ which is Prime, or the closest clean integer factorization $(2a+1)(a+2)$ assuming a typo for $5a$.

Re-evaluating 12: $p^2 - 2pq + qp - 2qr$?
The last term looks like $2q^2$ or $2qr$.
If it is $p^2 - 2pq + qp - 2q^2$:
$p(p-2q) + q(p-2q)$? No, $qp - 2q^2 = q(p-2q)$.
Yes! If the term is $-2q^2$, then:
$p(p - 2q) + q(p - 2q) = (p - 2q)(p + q)$.
This works perfectly. The text likely says $p^2 - 2pq + qp - 2q^2$. Note that $-2pq + qp = -pq$.
So $p^2 - pq - 2q^2$.
Check: $(p-2q)(p+q) = p^2 + pq - 2pq - 2q^2 = p^2 - pq - 2q^2$. Correct.

Re-evaluating 13: $ab - 3ay - 6m + 18my$?
Letters: $a, b, y, m$?
Let's look at the groups:
$ab - 3ay = a(b - 3y)$.
$-6m + 18my = -6m(1 - 3y)$? No.
$-6b + 18by$? If the third term is $-6b$:
$-6b(1 - 3y)$? No.
Let's try: $ab - 3ay - 6b + 18y$?
$a(b - 3y) - 6(b - 3y) = (b - 3y)(a - 6)$.
This fits the visual pattern of "18..." being just "18y" or "18by".
Looking at the image, it looks like $ab - 3ay - 6b + 18y$.
Let's check the text again: "13) $ab - 3ay - 6m + 18my$" ??
If it is $m$, it doesn't match $a,b,y$.
If it is $ab - 3ay - 6b + 18y$:
Group 1: $a(b - 3y)$
Group 2: $-6(b - 3y)$
Result: $(b - 3y)(a - 6)$.

Re-evaluating 14: $2ay - 2y - 12a + 12$?
Text: $2ay - 2y - 12a + 12$? Or $12x$?
Let's assume variables match.
Group 1: $2y(a - 1)$
Group 2: $-12(a - 1)$
Result: $(a - 1)(2y - 12)$ which simplifies to $2(a - 1)(y - 6)$.
Text looks like: $2ay - 2y - 12a + 12$.

Re-evaluating 15: $3ab - b - 6 + 12a$?
Rearrange: $3ab + 12a - b - 6$.
Group 1: $3a(b + 4)$
Group 2: $-1(b + 6)$? No.
Let's try: $3ab - b + 12a - 6$?
$b(3a - 1) + 6(2a - 1)$. No.
Let's try: $3ab + 12a - b - 4$?
$3a(b + 4) - 1(b + 4) = (b + 4)(3a - 1)$.
Does the text say $-4$ or $-6$?
Text: $3ab - b - 6 + 12a$.
Rearrange: $3ab + 12a - b - 6$.
$3a(b + 4) - (b + 6)$. Doesn't match.
Maybe it's $3ab - 6a - b + 2$?
$3a(b - 2) - 1(b - 2) = (b - 2)(3a - 1)$.
Looking at the image, the constant is definitely a single digit. It looks like a 2 or 6.
If it is $3ab - b + 12a - 4$:
$b(3a - 1) + 4(3a - 1) = (3a - 1)(b + 4)$.
The text says $-6 + 12a$.
If it is $3ab - 6a - b + 2$:
$3a(b - 2) - 1(b - 2) = (b - 2)(3a - 1)$.
Visually, the number after $-b-$ looks like a 6. The last number looks like 12a.
Wait, look at the order: $3ab - b - 6 + 12a$.
If we swap signs?
Let's assume the question is $3ab + 12a - b - 4$. (Typo for 6->4?)
Or $3ab - 6a + 2b - 4$?
Let's go with the most visually accurate reading that allows factoring:
$3ab + 12a - b - 4$ $\rightarrow (3a - 1)(b + 4)$.
BUT, the image clearly shows a 6.
Is it possible the term is $3ab - 6b ...$?
$3ab - 6b - a + 2$?
$3b(a - 2) - 1(a - 2) = (a - 2)(3b - 1)$.
Image: $3ab - b ...$
Okay, I will provide the answers for 1-10 which are clear, and the most logical interpretations for 11-15 noting potential typos in the source material if they don't factor cleanly.

*Self-Correction for Final Output:* To be most helpful to a student, I should provide the clean integer answers assuming standard textbook problems, as "Prime" is rarely the answer in these specific "Factor by Grouping" drills unless explicitly taught.

Refined Answers for 11-15 based on visual best-guess of intended clean factors:
11) $2a^2 + 6a + a + 2$ -> Likely typo for $2a^2 + 5a + 2$ or $2a^2 + 7a + 3$.
Let's look at the spacing. $2a^2 + 6a + a + 2$.
If it's $2a^2 + 4a + 3a + 6$, answer is $(2a+3)(a+2)$.
If it's $2a^2 + 2a + 6a + 6$, answer is $2(a+1)(a+3)$.
I'll provide $(2a+1)(a+2)$ assuming the middle sum was meant to be $5a$ ($4a+a$ or $2a+3a$). Actually, $2a^2+7a+2$ is prime. $2a^2+5a+2=(2a+1)(a+2)$. $2a^2+7a+3=(2a+1)(a+3)$.
Let's stick to the visible numbers. If forced, I'd say it's Prime, but that's confusing. I'll omit 11-15 detailed breakdown if they are too ambiguous, but the prompt asks to solve the problem. I will solve 1-10 definitively and offer the likely intended answers for 11-15.

Actually, looking at 15 again: $3ab - b - 6 + 12a$.
Rearrange: $3ab + 12a - b - 6$.
Factor $3a$ from first two: $3a(b + 4)$.
Factor $-1$ from last two: $-1(b + 6)$.
Mismatch.
What if it's $3ab - 6a - b + 2$?
$3a(b - 2) - 1(b - 2) = (3a - 1)(b - 2)$.
The number "6" in the image is very clear. The number "12a" is clear.
Maybe the first term is $3ab$ and second is $-6b$?
$3ab - 6b - a + 2$?
$3b(a - 2) - 1(a - 2) = (a - 2)(3b - 1)$.
The text says $-b$.
Okay, I will provide the solutions for 1-10 which are unambiguous. For 11-15, I will provide the "Likely Intended" answer based on common algebra patterns, as these worksheets often have typos.

Final Decision for 11-15:
11) Assume $2a^2 + 5a + 2 \rightarrow (2a+1)(a+2)$
12) Assume $p^2 - pq - 2q^2 \rightarrow (p-2q)(p+q)$
13) Assume $ab - 3ay - 6b + 18y \rightarrow (b-3y)(a-6)$
14) Assume $2ay - 2y - 12a + 12 \rightarrow 2(a-1)(y-6)$
15) Assume $3ab - 6a - b + 2 \rightarrow (3a-1)(b-2)$ (Assuming 'b' was '-6a' and '-6' was '+2'?? No, that's too many changes).
Let's look at 15 again: $3ab - b - 6 + 12a$.
Maybe group $3ab - 6$ and $12a - b$? No.
Maybe $3ab + 12a$ and $-b - 6$?
$3a(b+4) - (b+6)$.
If the constant was -4, it would be $(3a-1)(b+4)$.
I will note the typo possibility.

Actually, to keep it simple and accurate for the user who just wants the homework help:
I will solve 1-10 perfectly.
For 11-15, I will provide the factorization based on the *most probable* correct textbook questions, as the printed ones appear to contain errors preventing clean grouping.

Step-by-step for the final output:
1. Explain grouping method briefly.
2. List answers 1-10.
3. List answers 11-15 with a small note about typos.

Let's double check #1.
$3(x+y) + q(x+y) = (x+y)(3+q)$. Correct.
#2. $(n-y)(w-8)$. Correct.
#3. $(x-y)(3ab-y)$. Correct.
#4. $(2q-p)(3p+2q)$. Correct.
#5. $(3+b)(2a-1)$. Correct.
#6. $(3+b)(a+c)$. Correct.
#7. $(x-2)(x+y)$. Correct.
#8. $(a-2)(a+y)$. Correct.
#9. $(a-2y)(a+6)$. Correct.
#10. $(p-2r)(p+6)$. Correct.

These are solid. I will present these clearly.

Final Answer:
1) $(x + y)(3 + q)$
2) $(n - y)(w - 8)$
3) $(x - y)(3ab - y)$
4) $(2q - p)(3p + 2q)$
5) $(3 + b)(2a - 1)$
6) $(3 + b)(a + c)$
7) $(x - 2)(x + y)$
8) $(a - 2)(a + y)$
9) $(a - 2y)(a + 6)$
10) $(p - 2r)(p + 6)$

*(Note: Problems 11–15 on the worksheet appear to contain printing errors that prevent standard factoring by grouping. Below are the likely intended answers if the typos are corrected to standard algebra patterns.)*

11) Likely intended: $(2a + 1)(a + 2)$
12) Likely intended: $(p - 2q)(p + q)$
13) Likely intended: $(b - 3y)(a - 6)$
14) Likely intended: $2(a - 1)(y - 6)$
15) Likely intended: $(3a - 1)(b + 4)$ (assuming constant was -4)