Factoring Quadratic Expressions Worksheet | Fun and Engaging ... - Free Printable

Let’s solve each problem step by step. We’ll start with Section A, then Section B, and finally the Extension.

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Section A: Simplify then factorise

We need to first simplify (expand and combine like terms), then factorise the quadratic expression.

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1) x² – 6x – 2x + 12

Simplify:
x² – 8x + 12

Now factorise:
Find two numbers that multiply to 12 and add to -8 → -6 and -2
→ (x – 6)(x – 2)

✔ Final Answer for 1: (x – 6)(x – 2)

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2) d(d – 5) – 84

Expand:
d² – 5d – 84

Factorise:
Need two numbers that multiply to -84 and add to -5 → -12 and 7
→ (d – 12)(d + 7)

✔ Final Answer for 2: (d – 12)(d + 7)

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3) b² + 2(b – 4)

Expand:
b² + 2b – 8

Factorise:
Multiply to -8, add to 2 → 4 and -2
→ (b + 4)(b – 2)

✔ Final Answer for 3: (b + 4)(b – 2)

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4) x² – 3(2x + 9)

Expand:
x² – 6x – 27

Factorise:
Multiply to -27, add to -6 → -9 and 3
→ (x – 9)(x + 3)

✔ Final Answer for 4: (x – 9)(x + 3)

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5) c(c + 8) – 48

Expand:
c² + 8c – 48

Factorise:
Multiply to -48, add to 8 → 12 and -4
→ (c + 12)(c – 4)

✔ Final Answer for 5: (c + 12)(c – 4)

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6) 3a(a – 2) – 4a + 3

Expand:
3a² – 6a – 4a + 3 = 3a² – 10a + 3

Factorise:
Multiply 3×3=9, find factors of 9 that add to -10 → -9 and -1
Split middle term:
3a² – 9a – a + 3
Group: (3a² – 9a) + (-a + 3) = 3a(a – 3) –1(a – 3)
→ (3a – 1)(a – 3)

✔ Final Answer for 6: (3a – 1)(a – 3)

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7) 5w(w – 2) – 4w – 3

Expand:
5w² – 10w – 4w – 3 = 5w² – 14w – 3

Factorise:
Multiply 5×(-3)= -15, find factors of -15 that add to -14 → -15 and 1
Split: 5w² – 15w + w – 3
Group: 5w(w – 3) +1(w – 3)
→ (5w + 1)(w – 3)

✔ Final Answer for 7: (5w + 1)(w – 3)

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8) 3(6 – 5s) + s² + s²

Simplify:
18 – 15s + 2s² → rearrange: 2s² – 15s + 18

Factorise:
Multiply 2×18=36, factors of 36 that add to -15 → -12 and -3
Split: 2s² – 12s – 3s + 18
Group: 2s(s – 6) –3(s – 6)
→ (2s – 3)(s – 6)

✔ Final Answer for 8: (2s – 3)(s – 6)

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9) 3 + 2y(4y + 5)

Expand:
3 + 8y² + 10y → rearrange: 8y² + 10y + 3

Factorise:
Multiply 8×3=24, factors of 24 that add to 10 → 6 and 4
Split: 8y² + 6y + 4y + 3
Group: 2y(4y + 3) +1(4y + 3)
→ (2y + 1)(4y + 3)

✔ Final Answer for 9: (2y + 1)(4y + 3)

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10) 9x² – (x – 3)²

First expand (x – 3)² = x² – 6x + 9
So: 9x² – (x² – 6x + 9) = 9x² – x² + 6x – 9 = 8x² + 6x – 9

Factorise:
Multiply 8×(-9)= -72, factors of -72 that add to 6 → 12 and -6
Split: 8x² + 12x – 6x – 9
Group: 4x(2x + 3) –3(2x + 3)
→ (4x – 3)(2x + 3)

✔ Final Answer for 10: (4x – 3)(2x + 3)

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Section B: Factorise using difference of squares or other methods

Remember: Difference of squares: a² – b² = (a – b)(a + b)

Also look for common factors first!

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1) x² – 4 = (x – 2)(x + 2)

✔ Final Answer for 1: (x – 2)(x + 2)

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2) s² – 25 = (s – 5)(s + 5)

✔ Final Answer for 2: (s – 5)(s + 5)

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3) t² – 64 = (t – 8)(t + 8)

✔ Final Answer for 3: (t – 8)(t + 8)

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4) 9 – y² = (3 – y)(3 + y)

✔ Final Answer for 4: (3 – y)(3 + y)

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5) 49 – p² = (7 – p)(7 + p)

✔ Final Answer for 5: (7 – p)(7 + p)

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6) 4q² – 121 = (2q)² – 11² = (2q – 11)(2q + 11)

✔ Final Answer for 6: (2q – 11)(2q + 11)

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7) 81 – 25k² = 9² – (5k)² = (9 – 5k)(9 + 5k)

✔ Final Answer for 7: (9 – 5k)(9 + 5k)

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8) 1 – 400d² = 1² – (20d)² = (1 – 20d)(1 + 20d)

✔ Final Answer for 8: (1 – 20d)(1 + 20d)

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9) 600v² – 6

Factor out GCF: 6(100v² – 1)
Then 100v² – 1 = (10v)² – 1² = (10v – 1)(10v + 1)
→ 6(10v – 1)(10v + 1)

✔ Final Answer for 9: 6(10v – 1)(10v + 1)

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10) a² – b² = (a – b)(a + b)

✔ Final Answer for 10: (a – b)(a + b)

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11) x² – 9y² = x² – (3y)² = (x – 3y)(x + 3y)

✔ Final Answer for 11: (x – 3y)(x + 3y)

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12) 4c² – d² = (2c)² – d² = (2c – d)(2c + d)

✔ Final Answer for 12: (2c – d)(2c + d)

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13) 16s² – 9t² = (4s)² – (3t)² = (4s – 3t)(4s + 3t)

✔ Final Answer for 13: (4s – 3t)(4s + 3t)

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14) 49w² – 100v² = (7w)² – (10v)² = (7w – 10v)(7w + 10v)

✔ Final Answer for 14: (7w – 10v)(7w + 10v)

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15) 32p² – 18q²

GCF is 2: 2(16p² – 9q²)
Then 16p² – 9q² = (4p)² – (3q)² = (4p – 3q)(4p + 3q)
→ 2(4p – 3q)(4p + 3q)

✔ Final Answer for 15: 2(4p – 3q)(4p + 3q)

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16) 48x² – 12y²

GCF is 12: 12(4x² – y²)
Then 4x² – y² = (2x)² – y² = (2x – y)(2x + y)
→ 12(2x – y)(2x + y)

✔ Final Answer for 16: 12(2x – y)(2x + y)

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17) 45a² – 125b²

GCF is 5: 5(9a² – 25b²)
Then 9a² – 25b² = (3a)² – (5b)² = (3a – 5b)(3a + 5b)
→ 5(3a – 5b)(3a + 5b)

✔ Final Answer for 17: 5(3a – 5b)(3a + 5b)

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18) 72x² – 242y²

GCF is 2: 2(36x² – 121y²)
Then 36x² – 121y² = (6x)² – (11y)² = (6x – 11y)(6x + 11y)
→ 2(6x – 11y)(6x + 11y)

✔ Final Answer for 18: 2(6x – 11y)(6x + 11y)

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19) a²b² – c² = (ab)² – c² = (ab – c)(ab + c)

✔ Final Answer for 19: (ab – c)(ab + c)

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20) 9s – 4s³

Factor out GCF: s(9 – 4s²)
Then 9 – 4s² = 3² – (2s)² = (3 – 2s)(3 + 2s)
→ s(3 – 2s)(3 + 2s)

✔ Final Answer for 20: s(3 – 2s)(3 + 2s)

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21) (xy)² – 4z² = (xy)² – (2z)² = (xy – 2z)(xy + 2z)

✔ Final Answer for 21: (xy – 2z)(xy + 2z)

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22) 64t⁴ – 16s⁴

GCF is 16: 16(4t⁴ – s⁴)
Note: 4t⁴ – s⁴ = (2t²)² – (s²)² = (2t² – s²)(2t² + s²)
But 2t² – s² can be factored further? Only if it’s difference of squares — but 2t² isn’t a perfect square unless we use radicals, which we don’t want here. So stop at:

Wait — actually, 4t⁴ – s⁴ = (2t² – s²)(2t² + s²), and neither factor is difference of squares over integers. But let me check original:

Actually, 64t⁴ – 16s⁴ = 16(4t⁴ – s⁴) = 16[(2t²)² – (s²)²] = 16(2t² – s²)(2t² + s²)

But maybe they expect full factorisation? Let’s see if 4t⁴ – s⁴ can be written as difference of squares again? No, because 2t² and s² are not both squares unless t and s are specific.

Alternatively, perhaps write as:

64t⁴ – 16s⁴ = (8t²)² – (4s²)² = (8t² – 4s²)(8t² + 4s²) → then factor out 4 from each: 4(2t² – s²) * 4(2t² + s²) = 16(2t² – s²)(2t² + s²) — same as before.

So final answer: 16(2t² – s²)(2t² + s²)

But wait — 2t² – s² cannot be factored further with integer coefficients. So this is fine.

✔ Final Answer for 22: 16(2t² – s²)(2t² + s²)

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23) (4x²)² – 36y² = 16x⁴ – 36y²

This is difference of squares: (4x²)² – (6y)² = (4x² – 6y)(4x² + 6y)

But we can factor out 2 from each: 2(2x² – 3y) * 2(2x² + 3y) = 4(2x² – 3y)(2x² + 3y)

Wait — better to factor GCF first:

Original: (4x²)² – 36y² = 16x⁴ – 36y²

GCF of 16 and 36 is 4: 4(4x⁴ – 9y²)

Now 4x⁴ – 9y² = (2x²)² – (3y)² = (2x² – 3y)(2x² + 3y)

→ 4(2x² – 3y)(2x² + 3y)

✔ Final Answer for 23: 4(2x² – 3y)(2x² + 3y)

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24) 27a⁴ – 12b²

GCF is 3: 3(9a⁴ – 4b²)

Now 9a⁴ – 4b² = (3a²)² – (2b)² = (3a² – 2b)(3a² + 2b)

→ 3(3a² – 2b)(3a² + 2b)

✔ Final Answer for 24: 3(3a² – 2b)(3a² + 2b)

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Extension: Using difference of two squares

These are expressions in the form A² – B², so factor as (A – B)(A + B)

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1) 4x² – (x – 2)²

Write as: (2x)² – (x – 2)² → difference of squares

= [2x – (x – 2)][2x + (x – 2)]
= (2x – x + 2)(2x + x – 2)
= (x + 2)(3x – 2)

✔ Final Answer for Extension 1: (x + 2)(3x – 2)

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2) (2x + 1)² – (x – 4)²

Difference of squares: A = (2x+1), B = (x–4)

= [(2x + 1) – (x – 4)][(2x + 1) + (x – 4)]
= (2x + 1 – x + 4)(2x + 1 + x – 4)
= (x + 5)(3x – 3)

Now factor out 3 from second bracket: 3(x – 1)

→ (x + 5) * 3(x – 1) = 3(x + 5)(x – 1)

✔ Final Answer for Extension 2: 3(x + 5)(x – 1)

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## ✔ FINAL ANSWERS SUMMARY:

Section A:

1) (x – 6)(x – 2)
2) (d – 12)(d + 7)
3) (b + 4)(b – 2)
4) (x – 9)(x + 3)
5) (c + 12)(c – 4)
6) (3a – 1)(a – 3)
7) (5w + 1)(w – 3)
8) (2s – 3)(s – 6)
9) (2y + 1)(4y + 3)
10) (4x – 3)(2x + 3)

Section B:

1) (x – 2)(x + 2)
2) (s – 5)(s + 5)
3) (t – 8)(t + 8)
4) (3 – y)(3 + y)
5) (7 – p)(7 + p)
6) (2q – 11)(2q + 11)
7) (9 – 5k)(9 + 5k)
8) (1 – 20d)(1 + 20d)
9) 6(10v – 1)(10v + 1)
10) (a – b)(a + b)
11) (x – 3y)(x + 3y)
12) (2c – d)(2c + d)
13) (4s – 3t)(4s + 3t)
14) (7w – 10v)(7w + 10v)
15) 2(4p – 3q)(4p + 3q)
16) 12(2x – y)(2x + y)
17) 5(3a – 5b)(3a + 5b)
18) 2(6x – 11y)(6x + 11y)
19) (ab – c)(ab + c)
20) s(3 – 2s)(3 + 2s)
21) (xy – 2z)(xy + 2z)
22) 16(2t² – s²)(2t² + s²)
23) 4(2x² – 3y)(2x² + 3y)
24) 3(3a² – 2b)(3a² + 2b)

Extension:

1) (x + 2)(3x – 2)
2) 3(x + 5)(x – 1)

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Final Answer:
All answers listed above per question number.