SOLUTION: Factoring review worksheet 56 may 9 2022 10 08 40 am ... - Free Printable

Let’s go through each problem in Set 1 step by step. The goal is to factor each polynomial by finding the greatest common factor (GCF) — that means the biggest number and variable(s) that divide evenly into every term.

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Problem 1: 21x – 15y

- Look at coefficients: 21 and 15 → GCF is 3.
- Variables: x and y → no common variable.
- So, factor out 3:
→ 3(7x – 5y)

✔ Correct as written on paper.

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Problem 2: 14c² + 2c

- Coefficients: 14 and 2 → GCF is 2.
- Variables: c² and c → lowest power is c¹ → so factor out c.
- GCF = 2c
- Divide each term by 2c:
- 14c² ÷ 2c = 7c
- 2c ÷ 2c = 1
→ 2c(7c + 1)

✔ Correct as written.

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Problem 3: 5p² + 12q²

- Coefficients: 5 and 12 → GCF is 1 (no common factor other than 1).
- Variables: p² and q² → different variables → no common variable.
→ Cannot factor further → “prime” is correct.

✔ Correct.

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Problem 4: 8x³y⁴ – 22x⁵y⁶

- Coefficients: 8 and 22 → GCF is 2.
- Variables:
- x³ and x⁵ → lowest power is x³
- y⁴ and y⁶ → lowest power is y⁴
→ GCF = 2x³y⁴
- Divide each term:
- 8x³y⁴ ÷ 2x³y⁴ = 4
- 22x⁵y⁶ ÷ 2x³y⁴ = 11x²y²
→ 2x³y⁴(4 – 11x²y²)

⚠️ On paper it says: `2x³y⁴(4 - 11x²y²)` — but handwriting looks like “11x^a y^a” — probably meant x²y². Assuming typo in writing, answer is correct if exponents are 2.

✔ Correct with assumption.

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Problem 5: 10a²b² + 9ab² – a²b

This one is trickier because there are three terms. Let’s find GCF of all three.

Terms:
① 10a²b²
② 9ab²
③ –a²b

Look at coefficients: 10, 9, 1 → GCF = 1? Wait — let’s check again.

Actually, look for what divides ALL THREE:

- Coefficients: 10, 9, 1 → only common factor is 1.
- Variables:
- a², a, a² → lowest power of a is a¹
- b², b², b → lowest power of b is b¹
→ So GCF = ab

Now divide each term by ab:

① 10a²b² ÷ ab = 10ab
② 9ab² ÷ ab = 9b
③ –a²b ÷ ab = –a

So factored form:
→ ab(10ab + 9b – a)

But wait — on paper it says: `ab(10ab + 9b - a)` — which matches!

However, note: sometimes people write it as ab(10ab – a + 9b), same thing.

✔ Correct.

BUT — hold on! Let me double-check if we can factor more inside? No, because 10ab, 9b, –a have no common factor.

Also, original expression: 10a²b² + 9ab² – a²b — yes, factoring ab gives exactly that.

✔ Final answer: ab(10ab + 9b – a)

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Problem 6: 21m⁵n² + 6m³n + 15m²

Three terms again.

Coefficients: 21, 6, 15 → GCF is 3.

Variables:
- m⁵, m³, m² → lowest power is m²
- n², n, (no n in last term!) → so n is NOT common to all → cannot factor n.

Wait — third term is 15m² — has NO n. So n is not part of GCF.

→ GCF = 3m²

Divide each term by 3m²:

① 21m⁵n² ÷ 3m² = 7m³n²
② 6m³n ÷ 3m² = 2mn
③ 15m² ÷ 3m² = 5

→ 3m²(7m³n² + 2mn + 5)

On paper it says: `3m²(7m³n² + 2mn + 5)` — perfect.

✔ Correct.

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Now, just to be thorough — let’s verify Problem 5 again since it had three terms and could be tricky.

Original: 10a²b² + 9ab² – a²b

Factor out ab:

= ab*(10ab) + ab*(9b) + ab*(-a)
= ab(10ab + 9b – a)

Yes — correct.

Some might think to rearrange: ab(10ab – a + 9b) — also fine.

No further factoring possible.

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All answers on the sheet appear correct except possibly Problem 4 where handwriting may have miswritten exponents — but assuming intent was x²y², it's correct.

Final Answer:

1. 3(7x – 5y)
2. 2c(7c + 1)
3. prime
4. 2x³y⁴(4 – 11x²y²)
5. ab(10ab + 9b – a)
6. 3m²(7m³n² + 2mn + 5)