Factoring Quadratic Expressions Worksheet | Fun and Engaging ... - Free Printable
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Step-by-step solution for: Factoring Quadratic Expressions Worksheet | Fun and Engaging ...
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Show Answer Key & Explanations
Step-by-step solution for: Factoring Quadratic Expressions Worksheet | Fun and Engaging ...
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
Step 1: Combine like terms → x² – 8x + 12
Step 2: Factorise → Find two numbers that multiply to 12 and add to -8 → -6 and -2
→ (x – 6)(x – 2)
✔ Final: (x – 6)(x – 2)
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2) d(d – 5) – 84
Step 1: Expand → d² – 5d – 84
Step 2: Factorise → Two numbers that multiply to -84 and add to -5 → -12 and 7
→ (d – 12)(d + 7)
✔ Final: (d – 12)(d + 7)
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3) b² + 2(b – 4)
Step 1: Expand → b² + 2b – 8
Step 2: Factorise → Multiply to -8, add to 2 → 4 and -2
→ (b + 4)(b – 2)
✔ Final: (b + 4)(b – 2)
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4) x² – 3(2x + 9)
Step 1: Expand → x² – 6x – 27
Step 2: Factorise → Multiply to -27, add to -6 → -9 and 3
→ (x – 9)(x + 3)
✔ Final: (x – 9)(x + 3)
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5) c(c + 8) – 48
Step 1: Expand → c² + 8c – 48
Step 2: Factorise → Multiply to -48, add to 8 → 12 and -4
→ (c + 12)(c – 4)
✔ Final: (c + 12)(c – 4)
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6) 3a(a – 2) – 4a + 3
Step 1: Expand → 3a² – 6a – 4a + 3 = 3a² – 10a + 3
Step 2: Factorise → Use AC method: 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: (3a – 1)(a – 3)
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7) 5w(w – 2) – 4w – 3
Step 1: Expand → 5w² – 10w – 4w – 3 = 5w² – 14w – 3
Step 2: Factorise → AC = 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: (5w + 1)(w – 3)
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8) 3(6 – 5s) + s² + s²
Step 1: Simplify → 18 – 15s + 2s² → rewrite as 2s² – 15s + 18
Step 2: Factorise → AC = 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: (2s – 3)(s – 6)
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9) 3 + 2y(4y + 5)
Step 1: Expand → 3 + 8y² + 10y → 8y² + 10y + 3
Step 2: Factorise → AC = 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: (2y + 1)(4y + 3)
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10) 9x² – (x – 3)²
Step 1: Expand (x – 3)² → x² – 6x + 9
So: 9x² – (x² – 6x + 9) = 9x² – x² + 6x – 9 = 8x² + 6x – 9
Step 2: Factorise → AC = 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: (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)
2) s² – 25 → (s – 5)(s + 5)
3) t² – 64 → (t – 8)(t + 8)
4) 9 – y² → (3 – y)(3 + y)
5) 49 – p² → (7 – p)(7 + p)
6) 4q² – 121 → (2q – 11)(2q + 11)
7) 81 – 25k² → (9 – 5k)(9 + 5k)
8) 1 – 400d² → (1 – 20d)(1 + 20d)
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9) 600v² – 6
First, factor out GCF: 6 → 6(100v² – 1)
Now 100v² – 1 is difference of squares → (10v – 1)(10v + 1)
→ 6(10v – 1)(10v + 1)
✔ Final: 6(10v – 1)(10v + 1)
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10) a² – b² → (a – b)(a + b)
11) x² – 9y² → (x – 3y)(x + 3y)
12) 4c² – d² → (2c – d)(2c + d)
13) 16s² – 9t² → (4s – 3t)(4s + 3t)
14) 49w² – 100v² → (7w – 10v)(7w + 10v)
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15) 32p² – 18q²
GCF = 2 → 2(16p² – 9q²) → 2(4p – 3q)(4p + 3q)
✔ Final: 2(4p – 3q)(4p + 3q)
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16) 48x² – 12y²
GCF = 12 → 12(4x² – y²) → 12(2x – y)(2x + y)
✔ Final: 12(2x – y)(2x + y)
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17) 45a² – 125b²
GCF = 5 → 5(9a² – 25b²) → 5(3a – 5b)(3a + 5b)
✔ Final: 5(3a – 5b)(3a + 5b)
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18) 72x² – 242y²
GCF = 2 → 2(36x² – 121y²) → 2(6x – 11y)(6x + 11y)
✔ Final: 2(6x – 11y)(6x + 11y)
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19) a²b² – c² → (ab – c)(ab + c)
20) 9s – 4s³ → Factor out s: s(9 – 4s²) → s(3 – 2s)(3 + 2s)
21) (xy)² – 4z² → (xy – 2z)(xy + 2z)
22) 64t⁴ – 16s⁴ → GCF = 16 → 16(4t⁴ – s⁴)
But 4t⁴ – s = (2t²)² – (s²)² → (2t² – s²)(2t² + s²)
And 2t² – s² can be factored further? Only if it's difference of squares — but 2 isn’t square. So stop here? Wait — actually, 4t⁴ – s is also (2t² – s²)(2t² + s²), and 2t² – s² is not factorable over integers. But wait — original was 64t⁴ – 16s⁴ = 16(4t⁴ – s⁴). And 4t⁴ – s⁴ = (2t² – s²)(2t² + s²). But we can go further: 4t⁴ – s⁴ = (2t²)^2 - (s^2)^2 = (2t² - s²)(2t² + s²). Still no more integer factors. However, note: 64t⁴ – 16s⁴ = 16(4t⁴ – s⁴) = 16(2t² – s²)(2t² + s²). But actually, 4t⁴ – s⁴ can also be written as (2t² - s²)(2t² + s²), which is correct. Alternatively, think of it as (4t²)^2 - (4s²)^2? No, 64t⁴ = (8t²)^2, 16s⁴ = (4s²)^2 → so 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 thing. So final answer: 16(2t² – s²)(2t² + s²)
Wait — but 2t² – s² is not factorable unless we use radicals. So yes, this is fully factored over integers.
✔ Final: 16(2t² – s²)(2t² + s²)
Actually, let me double-check: 64t⁴ – 16s⁴ = 16(4t⁴ – s⁴). Now 4t⁴ – s⁴ = (2t²)^2 - (s²)^2 = (2t² - s²)(2t² + s²). Correct.
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23) (4x²)² – 36y² → 16x⁴ – 36y²
GCF = 4 → 4(4x⁴ – 9y²) → now 4x⁴ – 9y² = (2x²)^2 - (3y)^2 → (2x² – 3y)(2x² + 3y)
So overall: 4(2x² – 3y)(2x² + 3y)
✔ Final: 4(2x² – 3y)(2x² + 3y)
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24) 27a⁴ – 12b²
GCF = 3 → 3(9a⁴ – 4b²) → 9a⁴ – 4b² = (3a²)^2 - (2b)^2 → (3a² – 2b)(3a² + 2b)
So: 3(3a² – 2b)(3a² + 2b)
✔ Final: 3(3a² – 2b)(3a² + 2b)
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Extension: Using difference of two squares
These are in the form A² – B² = (A – B)(A + B)
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1) 4x² – (x – 2)²
This is [2x]² – [x – 2]² → difference of squares
→ (2x – (x – 2))(2x + (x – 2))
Simplify:
First bracket: 2x – x + 2 = x + 2
Second bracket: 2x + x – 2 = 3x – 2
→ (x + 2)(3x – 2)
✔ Final: (x + 2)(3x – 2)
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2) (2x + 1)² – (x – 4)²
Difference of squares: A = 2x+1, B = x–4
→ (A – B)(A + B)
A – B = (2x + 1) – (x – 4) = 2x + 1 – x + 4 = x + 5
A + B = (2x + 1) + (x – 4) = 3x – 3
→ (x + 5)(3x – 3)
Factor out 3 from second bracket: 3(x – 1)
So: 3(x + 5)(x – 1)
✔ Final: 3(x + 5)(x – 1)
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Final Answer:
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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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
Step 1: Combine like terms → x² – 8x + 12
Step 2: Factorise → Find two numbers that multiply to 12 and add to -8 → -6 and -2
→ (x – 6)(x – 2)
✔ Final: (x – 6)(x – 2)
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2) d(d – 5) – 84
Step 1: Expand → d² – 5d – 84
Step 2: Factorise → Two numbers that multiply to -84 and add to -5 → -12 and 7
→ (d – 12)(d + 7)
✔ Final: (d – 12)(d + 7)
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3) b² + 2(b – 4)
Step 1: Expand → b² + 2b – 8
Step 2: Factorise → Multiply to -8, add to 2 → 4 and -2
→ (b + 4)(b – 2)
✔ Final: (b + 4)(b – 2)
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4) x² – 3(2x + 9)
Step 1: Expand → x² – 6x – 27
Step 2: Factorise → Multiply to -27, add to -6 → -9 and 3
→ (x – 9)(x + 3)
✔ Final: (x – 9)(x + 3)
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5) c(c + 8) – 48
Step 1: Expand → c² + 8c – 48
Step 2: Factorise → Multiply to -48, add to 8 → 12 and -4
→ (c + 12)(c – 4)
✔ Final: (c + 12)(c – 4)
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6) 3a(a – 2) – 4a + 3
Step 1: Expand → 3a² – 6a – 4a + 3 = 3a² – 10a + 3
Step 2: Factorise → Use AC method: 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: (3a – 1)(a – 3)
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7) 5w(w – 2) – 4w – 3
Step 1: Expand → 5w² – 10w – 4w – 3 = 5w² – 14w – 3
Step 2: Factorise → AC = 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: (5w + 1)(w – 3)
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8) 3(6 – 5s) + s² + s²
Step 1: Simplify → 18 – 15s + 2s² → rewrite as 2s² – 15s + 18
Step 2: Factorise → AC = 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: (2s – 3)(s – 6)
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9) 3 + 2y(4y + 5)
Step 1: Expand → 3 + 8y² + 10y → 8y² + 10y + 3
Step 2: Factorise → AC = 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: (2y + 1)(4y + 3)
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10) 9x² – (x – 3)²
Step 1: Expand (x – 3)² → x² – 6x + 9
So: 9x² – (x² – 6x + 9) = 9x² – x² + 6x – 9 = 8x² + 6x – 9
Step 2: Factorise → AC = 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: (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)
2) s² – 25 → (s – 5)(s + 5)
3) t² – 64 → (t – 8)(t + 8)
4) 9 – y² → (3 – y)(3 + y)
5) 49 – p² → (7 – p)(7 + p)
6) 4q² – 121 → (2q – 11)(2q + 11)
7) 81 – 25k² → (9 – 5k)(9 + 5k)
8) 1 – 400d² → (1 – 20d)(1 + 20d)
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9) 600v² – 6
First, factor out GCF: 6 → 6(100v² – 1)
Now 100v² – 1 is difference of squares → (10v – 1)(10v + 1)
→ 6(10v – 1)(10v + 1)
✔ Final: 6(10v – 1)(10v + 1)
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10) a² – b² → (a – b)(a + b)
11) x² – 9y² → (x – 3y)(x + 3y)
12) 4c² – d² → (2c – d)(2c + d)
13) 16s² – 9t² → (4s – 3t)(4s + 3t)
14) 49w² – 100v² → (7w – 10v)(7w + 10v)
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15) 32p² – 18q²
GCF = 2 → 2(16p² – 9q²) → 2(4p – 3q)(4p + 3q)
✔ Final: 2(4p – 3q)(4p + 3q)
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16) 48x² – 12y²
GCF = 12 → 12(4x² – y²) → 12(2x – y)(2x + y)
✔ Final: 12(2x – y)(2x + y)
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17) 45a² – 125b²
GCF = 5 → 5(9a² – 25b²) → 5(3a – 5b)(3a + 5b)
✔ Final: 5(3a – 5b)(3a + 5b)
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18) 72x² – 242y²
GCF = 2 → 2(36x² – 121y²) → 2(6x – 11y)(6x + 11y)
✔ Final: 2(6x – 11y)(6x + 11y)
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19) a²b² – c² → (ab – c)(ab + c)
20) 9s – 4s³ → Factor out s: s(9 – 4s²) → s(3 – 2s)(3 + 2s)
21) (xy)² – 4z² → (xy – 2z)(xy + 2z)
22) 64t⁴ – 16s⁴ → GCF = 16 → 16(4t⁴ – s⁴)
But 4t⁴ – s = (2t²)² – (s²)² → (2t² – s²)(2t² + s²)
And 2t² – s² can be factored further? Only if it's difference of squares — but 2 isn’t square. So stop here? Wait — actually, 4t⁴ – s is also (2t² – s²)(2t² + s²), and 2t² – s² is not factorable over integers. But wait — original was 64t⁴ – 16s⁴ = 16(4t⁴ – s⁴). And 4t⁴ – s⁴ = (2t² – s²)(2t² + s²). But we can go further: 4t⁴ – s⁴ = (2t²)^2 - (s^2)^2 = (2t² - s²)(2t² + s²). Still no more integer factors. However, note: 64t⁴ – 16s⁴ = 16(4t⁴ – s⁴) = 16(2t² – s²)(2t² + s²). But actually, 4t⁴ – s⁴ can also be written as (2t² - s²)(2t² + s²), which is correct. Alternatively, think of it as (4t²)^2 - (4s²)^2? No, 64t⁴ = (8t²)^2, 16s⁴ = (4s²)^2 → so 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 thing. So final answer: 16(2t² – s²)(2t² + s²)
Wait — but 2t² – s² is not factorable unless we use radicals. So yes, this is fully factored over integers.
✔ Final: 16(2t² – s²)(2t² + s²)
Actually, let me double-check: 64t⁴ – 16s⁴ = 16(4t⁴ – s⁴). Now 4t⁴ – s⁴ = (2t²)^2 - (s²)^2 = (2t² - s²)(2t² + s²). Correct.
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23) (4x²)² – 36y² → 16x⁴ – 36y²
GCF = 4 → 4(4x⁴ – 9y²) → now 4x⁴ – 9y² = (2x²)^2 - (3y)^2 → (2x² – 3y)(2x² + 3y)
So overall: 4(2x² – 3y)(2x² + 3y)
✔ Final: 4(2x² – 3y)(2x² + 3y)
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24) 27a⁴ – 12b²
GCF = 3 → 3(9a⁴ – 4b²) → 9a⁴ – 4b² = (3a²)^2 - (2b)^2 → (3a² – 2b)(3a² + 2b)
So: 3(3a² – 2b)(3a² + 2b)
✔ Final: 3(3a² – 2b)(3a² + 2b)
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Extension: Using difference of two squares
These are in the form A² – B² = (A – B)(A + B)
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1) 4x² – (x – 2)²
This is [2x]² – [x – 2]² → difference of squares
→ (2x – (x – 2))(2x + (x – 2))
Simplify:
First bracket: 2x – x + 2 = x + 2
Second bracket: 2x + x – 2 = 3x – 2
→ (x + 2)(3x – 2)
✔ Final: (x + 2)(3x – 2)
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2) (2x + 1)² – (x – 4)²
Difference of squares: A = 2x+1, B = x–4
→ (A – B)(A + B)
A – B = (2x + 1) – (x – 4) = 2x + 1 – x + 4 = x + 5
A + B = (2x + 1) + (x – 4) = 3x – 3
→ (x + 5)(3x – 3)
Factor out 3 from second bracket: 3(x – 1)
So: 3(x + 5)(x – 1)
✔ Final: 3(x + 5)(x – 1)
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Final Answer:
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)
Parent Tip: Review the logic above to help your child master the concept of factoring quadratics worksheet a 1.