Factorising Trinomials — Solutions

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1. Factorise monic trinomials

   1. x² + 5x + 6: (x + 2)(x + 3)   [2×3=6, 2+3=5]
   2. x² + 8x + 15: (x + 3)(x + 5)   [3×5=15, 3+5=8]
   3. x² − 6x + 8: (x − 2)(x − 4)   [(−2)×(−4)=8, −2+(−4)=−6]
   4. x² − 9x + 14: (x − 2)(x − 7)   [(−2)×(−7)=14, −2+(−7)=−9]
   5. x² + 3x − 10: (x + 5)(x − 2)   [5×(−2)=−10, 5+(−2)=3]
   6. x² − 2x − 15: (x − 5)(x + 3)   [(−5)×3=−15, −5+3=−2]
   7. x² + 11x + 30: (x + 5)(x + 6)   [5×6=30, 5+6=11]
   8. x² − 7x + 12: (x − 3)(x − 4)   [(−3)×(−4)=12, −3+(−4)=−7]
   9. x² + x − 12: (x + 4)(x − 3)   [4×(−3)=−12, 4+(−3)=1]
   10. x² − 4x − 21: (x − 7)(x + 3)   [(−7)×3=−21, −7+3=−4]

2. Difference of two squares

   1. x² − 25: (x + 5)(x − 5)
   2. y² − 49: (y + 7)(y − 7)
   3. 4a² − 9: (2a + 3)(2a − 3)
   4. 9m² − 16: (3m + 4)(3m − 4)
   5. 25p² − 64: (5p + 8)(5p − 8)
   6. x² − 1: (x + 1)(x − 1)

3. Trinomials with HCF first or leading negative

   1. 2x² + 10x + 12: 2(x² + 5x + 6) = 2(x + 2)(x + 3)
   2. 3x² − 18x + 27: 3(x² − 6x + 9) = 3(x − 3)²
   3. −x² − 5x − 6: −1(x² + 5x + 6) = −(x + 2)(x + 3)
   4. −x² + 4x + 12: −1(x² − 4x − 12) = −(x − 6)(x + 2)
   5. Verify part (a): 2(x+2)(x+3) = 2(x²+5x+6) = 2x²+10x+12 ✓

4. Solving quadratics and area problems

   1. Solve x² + 5x + 6 = 0:
      1. Factorised: (x + 2)(x + 3) = 0
      2. Each factor = 0: x + 2 = 0 → x = −2;   x + 3 = 0 → x = −3
      3. Solutions: x = −2 or x = −3
   2. Solve x² − 4x − 12 = 0: (x − 6)(x + 2) = 0  →  x = 6 or x = −2
   3. Rectangle area x² + 9x + 20: (x + 4)(x + 5) cm. When x=3: length = 7 cm, width = 8 cm, area = 56 cm². Check: 9+27+20 = 56 ✓
   4. Square area (x² − 36): (x + 6)(x − 6). Side length = √(x²−36). For a square, all sides equal, so area must equal side². This expression is a difference of squares — not a perfect square in general. The area is positive when x > 6 (assuming x > 0).
   5. Two integers differing by 4 with product 77: x(x + 4) = 77 → x² + 4x − 77 = 0 → (x + 11)(x − 7) = 0 → x = 7 (taking positive value). The integers are 7 and 11. Check: 7×11 = 77 ✓

5. Factorising non-monic trinomials

   1. 2x² + 7x + 3: a×c = 6; find 6×1=6, 6+1=7 ✓. Rewrite: 2x²+6x+x+3 = 2x(x+3)+(x+3) = (2x+1)(x+3)
   2. 3x² + 10x + 8: a×c = 24; find 6×4=24, 6+4=10 ✓. Rewrite: 3x²+6x+4x+8 = 3x(x+2)+4(x+2) = (3x+4)(x+2)
   3. 2x² − 5x + 3: a×c = 6; find −3×−2=6, −3+(−2)=−5 ✓. Rewrite: 2x²−3x−2x+3 = x(2x−3)−1(2x−3) = (x−1)(2x−3)
   4. 5x² + 11x + 2: a×c = 10; find 10×1=10, 10+1=11 ✓. Rewrite: 5x²+10x+x+2 = 5x(x+2)+(x+2) = (5x+1)(x+2)
   5. 4x² − 4x − 3: a×c = −12; find −6×2=−12, −6+2=−4 ✓. Rewrite: 4x²−6x+2x−3 = 2x(2x−3)+(2x−3) = (2x+1)(2x−3)
   6. 6x² + x − 2: a×c = −12; find 4×−3=−12, 4+(−3)=1 ✓. Rewrite: 6x²+4x−3x−2 = 2x(3x+2)−1(3x+2) = (2x−1)(3x+2)

6. Mixed full factorisation

   1. 4x² + 8x − 12: HCF = 4. 4(x²+2x−3). Find 3×−1=−3, 3+(−1)=2. 4(x+3)(x−1)
   2. 6x² − 54: HCF = 6. 6(x²−9). Difference of squares: 6(x+3)(x−3)
   3. 5x² + 15x − 20: HCF = 5. 5(x²+3x−4). Find 4×−1=−4, 4+(−1)=3. 5(x+4)(x−1)
   4. 3x² − 12x + 9: HCF = 3. 3(x²−4x+3). Find −3×−1=3, −3+(−1)=−4. 3(x−3)(x−1)
   5. 2x² + 16x + 24: HCF = 2. 2(x²+8x+12). Find 6×2=12, 6+2=8. 2(x+6)(x+2)
   6. 10x² − 40: HCF = 10. 10(x²−4). Difference of squares: 10(x+2)(x−2)

7. Solving non-monic quadratics

   1. 2x² + 7x + 3 = 0: (2x+1)(x+3)=0 → x = −½ or x = −3
   2. 3x² − 8x + 4 = 0: a×c=12; −6×−2=12, −6+(−2)=−8. (3x−2)(x−2)=0 → x = ⅔ or x = 2
   3. 4x² − 4x − 3 = 0: (2x+1)(2x−3)=0 → x = −½ or x = &frac32;
   4. 2x² + 3x − 20 = 0: a×c=−40; 8×−5=−40, 8+(−5)=3. (2x−5)(x+4)=0 → x = &frac52; or x = −4
   5. 6x² − 7x − 3 = 0: a×c=−18; −9×2=−18, −9+2=−7. (3x+1)(2x−3)=0 → x = −⅓ or x = &frac32;

8. Quadratic equations from word problems

   1. Garden 2x²+9x+10: a×c=20; 4×5=20, 4+5=9. (2x+5)(x+2) m. When x=2: length=9 m, width=4 m, area=36 m². Check: 2(4)+9(2)+10=36 ✓
   2. Photo frame 3x²+13x+4: a×c=12; 12×1=12, 12+1=13. (3x+1)(x+4) ✓ matches dimensions. When x=5: (16)(9)=144 cm²; check 3(25)+13(5)+4=75+65+4=144 ✓
   3. h = −5t² + 15t = 0: −5t(t−3)=0 → t=0 or t=3. t=0 means the ball is at ground level at launch; t=3 means the ball returns to ground level after 3 seconds.

9. Difference of two squares applications

   1. 49² − 51²: (49+51)(49−51) = 100 × (−2) = −200
   2. n²−1 proof: n²−1 = (n+1)(n−1) by DOTS. For consecutive even integers 2k and 2k+2: product = 2k(2k+2) = 4k(k+1). Since k(k+1) is the product of two consecutive integers, one is even, so 2|k(k+1), giving 4×2=8 as a factor.
   3. 16x⁴ − 81y⁴: = (4x²+9y²)(4x²−9y²) = (4x²+9y²)(2x+3y)(2x−3y)
   4. x² − 36 = 0: (x+6)(x−6)=0 → x=6 or x=−6. Two solutions because both 6 and −6 square to 36.

10. Challenge: mixed extended problems

    1. Consecutive integers product 56: x(x+1)=56 → x²+x−56=0 → (x+8)(x−7)=0 → x=7 (positive). Integers: 7 and 8. Check: 7×8=56 ✓
    2. Triangular sail, area 24: ½(x+5)(x+2)=24 → (x+5)(x+2)=48 → x²+7x+10=48 → x²+7x−38=0 → (x+19)(x−2) → we check: try integer roots — (x−2)(x+19)=0 → x=2 (taking positive). Base=7 m, height=4 m. Area=½×7×4=14 ≠ 24... Let’s verify: ½(2+5)(2+2)=½×7×4=14. Recheck: x²+7x−38=0; discriminant=49+152=201 — not a perfect square. Try x=3: ½×8×5=20; x=4: ½×9×6=27. No integer solution; students should note x is between 3 and 4. For this problem x ≈ 3.6 (students may verify with a calculator).
    3. Values of k for x²+kx+9 to factorise: Need p×q=9 and p+q=k. Integer factor pairs of 9: (1,9),(3,3),(9,1),(−1,−9),(−3,−3),(−9,−1). Sums: 10, 6, 10, −10, −6, −10. k ∈ {−10, −6, 6, 10}
    4. Algebraic identity: (x+a)²−a² = x²+2ax+a²−a² = x²+2ax ✓. Therefore x²+2ax = x(x+2a), taking out the common factor x.