Answer:

【Answer】: 1. x = -2/3, x = -5, x = 5; 2. x = -2, x = 2, x = 3/2; 3. x = -4, x = -2/3, x = 1, x = 2; 4. x = -1, x = 1, x = -i√2, x = i√2, x = 2; 5. x = -1, x = 5, x = -3i, x = 3i; 6. x = -1, x = 1/3 (1 - 2/(9√193 - 125)^(1/3) + (9√193 - 125)^(1/3)), x = 1/6 (2 + (2 - 2i√3)/(9√193 - 125)^(1/3) + (-1 - i√3) (9√193 - 125)^(1/3)), x = 1/6 (2 + (2 + 2i√3)/(9√193 - 125)^(1/3) + i(√3 + i) (9√193 - 125)^(1/3)); 7. x = -3, x = -3 - √5, x = √5 - 3; 8. x = 1/2, x = 2 - √(7/2), x = 2 + √(7/2).

【Explanation】: 1. The equation (3x+2)(x^2-25)=0 can be solved by setting each factor equal to zero, resulting in the roots x = -2/3, x = -5, and x = 5. 2. The cubic equation 2x^3-3x^2-8x+12=0 has the roots x = -2, x = 2, and x = 3/2. 3. The quartic equation 3x^4+5x^3-28x^2+4x+16=0 has the roots x = -4, x = -2/3, x = 1, and x = 2. 4. The quintic equation x^5-2x^4+x^3-2x^2-2x+4=0 has five roots: x = -1, x = 1, x = -i√2, x = i√2, and x = 2. 5. The quartic equation x^4-4x^3+4x^2-36x-45=0 has the roots x = -1, x = 5, x = -3i, and x = 3i. 6. The quartic equation x^4+10x+9=0 has complex roots, and they are x = -1, x = 1/3 (1 - 2/(9√193 - 125)^(1/3) + (9√193 - 125)^(1/3)), x = 1/6 (2 + (2 - 2i√3)/(9√193 - 125)^(1/3) + (-1 - i√3) (9√193 - 125)^(1/3)), and x = 1/6 (2 + (2 + 2i√3)/(9√193 - 125)^(1/3) + i(√3 + i) (9√193 - 125)^(1/3)). 7. The cubic equation x^3+9x^2+22x+12=0 has the roots x = -3, x = -3 - √5, and x = √5 - 3. 8. The cubic equation 4x^3-18x^2+10x-1=0 has the roots x = 1/2, x = 2 - √(7/2), and x = 2 + √(7/2).

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