equations: q) \( x^{3}-3 x^{2}+2 x=0 \) r) \( 3 x^{3}-2 x^{2}-8 x=0 \) s) \( 6 x^{2}+7 x=-2 \) t) \( 9 x^{2}+24 x+15=0 \) u) \( 84 x^{3}+50 x^{2}+6 x=0 \) v) \( 2(x+1)^{2}=(2 x+2)(2 x+3) \)

We will solve each equation one by one. ──────────────────────────── q) Solve x³ – 3x² + 2x = 0. Step 1. Factor out x:   x (x² – 3x + 2) = 0 Step 2. Factor the quadratic:   x² – 3x + 2 = (x – 1)(x – 2) Thus, the equation factors as:   x (x – 1)(x – 2) = 0 Step 3. Set each factor equal to zero:   x = 0  or x – 1 = 0  or x – 2 = 0 Therefore, the solutions are:   x = 0, 1, 2 ──────────────────────────── r) Solve 3x³ – 2x² – 8x = 0. Step 1. Factor out x:   x (3x² – 2x – 8) = 0 So, one solution is:   x = 0 Step 2. Solve the quadratic 3x² – 2x – 8 = 0. Use the quadratic formula:   x = [–b ± √(b² – 4ac)] / (2a) Here, a = 3, b = –2, c = –8. Compute the discriminant:   Δ = (–2)² – 4·3·(–8) = 4 + 96 = 100 Then:   x = [2 ± 10] / (6) This gives:   x = (2 + 10) / 6 = 12/6 = 2   x = (2 – 10) / 6 = –8/6 = –4/3 Thus, the solutions are:   x = 0, 2, –4/3 ──────────────────────────── s) Solve 6x² + 7x = –2. Step 1. Bring all terms to one side:   6x² + 7x + 2 = 0 Step 2. Factor the quadratic. Look for two numbers that multiply to 6×2 = 12 and add to 7. They are 3 and 4. Write:   6x² + 3x + 4x + 2 = 0 Group terms:   (6x² + 3x) + (4x + 2) = 0 Factor:   3x(2x + 1) + 2(2x + 1) = 0 Factor common term (2x + 1):   (2x + 1)(3x + 2) = 0 Set each factor equal to zero:   2x + 1 = 0  ⟹  x = –1/2   3x + 2 = 0  ⟹  x = –2/3 Thus, the solutions are:   x = –1/2, –2/3 ──────────────────────────── t) Solve 9x² + 24x + 15 = 0. Step 1. Use the quadratic formula with a = 9, b = 24, c = 15. Compute the discriminant:   Δ = 24² – 4·9·15 = 576 – 540 = 36 Take the square root:   √36 = 6 Then:   x = [–24 ± 6] / (2·9) = [–24 ± 6] / 18 Thus,   x = (–24 + 6)/18 = –18/18 = –1   x = (–24 – 6)/18 = –30/18 = –5/3 So, the solutions are:   x = –1, –5/3 ──────────────────────────── u) Solve 84x³ + 50x² + 6x = 0. Step 1. Factor out the common factor 2x:   2x (42x² + 25x + 3) = 0 So one solution is:   x = 0 Step 2. Solve the quadratic 42x² + 25x + 3 = 0. Use the quadratic formula where a = 42, b = 25, c = 3. First, find the discriminant:   Δ = 25² – 4·42·3 = 625 – 504 = 121 √121 = 11 Now,   x = [–25 ± 11] / (2·42) = [–25 ± 11] / 84 This gives:   x = (–25 + 11) / 84 = –14/84 = –1/6   x = (–25 – 11) / 84 = –36/84 = –3/7 Thus, the solutions are:   x = 0, –1/6, –3/7 ──────────────────────────── v) Solve 2(x + 1)² = (2x + 2)(2x + 3). Step 1. Notice that 2x + 2 = 2(x + 1). Then the right-hand side becomes:   (2(x + 1))(2x + 3) The equation now reads:   2(x + 1)² = 2(x + 1)(2x + 3) Step 2. Divide both sides by 2 (provided 2 ≠ 0):   (x + 1)² = (x + 1)(2x + 3) Step 3. Subtract the right side from both sides:   (x + 1)² – (x + 1)(2x + 3) = 0 Factor out (x + 1):   (x + 1)[(x + 1) – (2x + 3)] = 0 Simplify the bracket:   (x + 1) – (2x + 3) = x + 1 – 2x – 3 = –x – 2 Thus, we have:   (x + 1)(–x – 2) = 0 Set each factor equal to zero: Factor 1: x + 1 = 0 ⟹ x = –1 Factor 2: –x – 2 = 0 ⟹ –x = 2 ⟹ x = –2 So, the solutions are:   x = –1, –2 ──────────────────────────── Summary of Answers: q) x = 0, 1, 2 r) x = 0, 2, –4/3 s) x = –1/2, –2/3 t) x = –1, –5/3 u) x = 0, –1/6, –3/7 v) x = –1, –2