Factor the following algebraic expressions: Q1 to Q10.

Let’s solve each problem one by one, step by step. We’ll factorize each expression carefully.

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Q1. 9a²b⁴ + 12a³b³

Step 1: Find the greatest common factor (GCF) of the coefficients:
9 and 12 → GCF is 3.

Step 2: For variables:
- a² and a³ → lowest power is a²
- b⁴ and b³ → lowest power is b³
So GCF of variables is a²b³

Step 3: Factor out 3a²b³:
= 3a²b³(3b + 4a)

✔ Final Answer for Q1: 3a²b³(4a + 3b)
*(We write 4a first to follow alphabetical order — not required, but neat)*

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Q2. a – 1 – (a – 1)² + ax – x

Step 1: Group terms that look similar. Notice “a – 1” appears twice, and “ax – x” can be factored.

Rewrite:
= (a – 1) – (a – 1)² + x(a – 1) [since ax – x = x(a – 1)]

Now all three terms have (a – 1):

Factor out (a – 1):
= (a – 1)[1 – (a – 1) + x]

Simplify inside the brackets:
1 – (a – 1) + x = 1 – a + 1 + x = (2 – a + x)

So:
= (a – 1)(x – a + 2)

✔ Final Answer for Q2: (a – 1)(x – a + 2)

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Q3. 25a(x – y) + 30b(y – x)

Note: (y – x) = –(x – y)

So rewrite second term:
= 25a(x – y) – 30b(x – y)

Now both terms have (x – y):

Factor out (x – y):
= (x – y)(25a – 30b)

Now factor out GCF from (25a – 30b): GCF of 25 and 30 is 5
= (x – y)·5(5a – 6b)

✔ Final Answer for Q3: 5(x – y)(5a – 6b)

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Q4. 6xy² + 9x²y – 21xy

Step 1: Coefficients: 6, 9, 21 → GCF is 3
Variables: each term has at least one x and one y → GCF is xy

So factor out 3xy:

= 3xy(2y + 3x – 7)

Check:
3xy × 2y = 6xy² ✔️
3xy × 3x = 9x²y ✔️
3xy × (–7) = –21xy ✔️

✔ Final Answer for Q4: 3xy(3x + 2y – 7)
*(Write in alphabetical order: 3x + 2y – 7)*

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Q5. 4x² – 25

This is a difference of squares:
A² – B² = (A – B)(A + B)

Here:
4x² = (2x)²
25 = 5²

So:
= (2x – 5)(2x + 5)

✔ Final Answer for Q5: (2x – 5)(2x + 5)

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Q6. 2x⁴ – 32

Step 1: Factor out GCF: 2
= 2(x⁴ – 16)

Now x⁴ – 16 is also a difference of squares:
x⁴ = (x²)², 16 = 4²
→ (x² – 4)(x² + 4)

But x² – 4 is ALSO a difference of squares:
= (x – 2)(x + 2)

So putting it all together:
= 2(x² + 4)(x – 2)(x + 2)

✔ Final Answer for Q6: 2(x – 2)(x + 2)(x² + 4)

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Q7. 25(2a – b)² – 81b²

This is a difference of squares again!

Let A = 5(2a – b), because 25(2a – b)² = [5(2a – b)]²
Let B = 9b, because 81b² = (9b)²

So:
= [5(2a – b)]² – (9b)²
= [5(2a – b) – 9b][5(2a – b) + 9b]

Now simplify each bracket:

First bracket:
5(2a – b) – 9b = 10a – 5b – 9b = 10a – 14b

Second bracket:
5(2a – b) + 9b = 10a – 5b + 9b = 10a + 4b

Now factor out common factors from each:

10a – 14b = 2(5a – 7b)
10a + 4b = 2(5a + 2b)

So overall:
= 2(5a – 7b) · 2(5a + 2b) = 4(5a – 7b)(5a + 2b)

✔ Final Answer for Q7: 4(5a – 7b)(5a + 2b)

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Q8. x⁸ – y⁸

Difference of squares multiple times.

x⁸ – y⁸ = (x⁴)² – (y⁴)² = (x⁴ – y)(x⁴ + y⁴)

Now x⁴ – y⁴ = (x²)² – (y²)² = (x² – y²)(x² + y²)

And x² – y² = (x – y)(x + y)

So full factorization:

= (x – y)(x + y)(x² + y²)(x⁴ + y⁴)

Note: x⁴ + y cannot be factored further over real numbers (it doesn’t break into real linear or quadratic factors easily without complex numbers).

✔ Final Answer for Q8: (x – y)(x + y)(x² + y²)(x⁴ + y)

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Q9. x² + y² – z² – 2xy

Rearrange terms:
= (x² – 2xy + y²) – z²

Notice x² – 2xy + y² = (x – y)²

So:
= (x – y)² – z² → difference of squares!

= [(x – y) – z][(x – y) + z]
= (x – y – z)(x – y + z)

✔ Final Answer for Q9: (x – y – z)(x – y + z)

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Q10. x² + 2xy + y² – z²

Group first three terms:
= (x² + 2xy + y²) – z²
= (x + y)² – z² → difference of squares!

= [(x + y) – z][(x + y) + z]
= (x + y – z)(x + y + z)

✔ Final Answer for Q10: (x + y – z)(x + y + z)

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Final Answers:

Q1: 3a²b³(4a + 3b)
Q2: (a – 1)(x – a + 2)
Q3: 5(x – y)(5a – 6b)
Q4: 3xy(3x + 2y – 7)
Q5: (2x – 5)(2x + 5)
Q6: 2(x – 2)(x + 2)(x² + 4)
Q7: 4(5a – 7b)(5a + 2b)
Q8: (x – y)(x + y)(x² + y²)(x⁴ + y)
Q9: (x – y – z)(x – y + z)
Q10: (x + y – z)(x + y + z)