Question

6x^4+13x^3-32x^2-45x+18因式分解

Answer 1

6x^4+13x^3-32x^2-45x+18
= 6x^4+18x^3-5x^3-15x^2-17x^2-51x+6x+18
= 6x^3(x+3) - 5x^2(x+3) - 17x(x+3) + 6(x+3)
= (x+3)(6x^3-5x^2-17x+6)
= (x+3)(6x^3-12x^2+7x^2-14x-3x+6)
= (x+3){6x^2(x-2)+7x(x-2)-3(x-2)}
= (x+3)(x-2)(6x^2+7x-3)
= (x+3)(x-2)(2x+3)(3x-1)

Answer 2

观察知x=2是方程f(x)=6x^4+13x^3-32x^2-45x+18=0的根。利用多项式除法，f(x)/(x-2)得商式
6x^3+25x^2+18x-9，余式为0。再则亮观察x=-3是商式的根，再利用多项式除法，带丛(6x^3+25x^2+18x-9)/(x+3)，得商式为6x^2+7x-3，余式为0。十字交叉法得孙行宽6x^2+7x-3=(2x+3)(3x-1)。
故6x^4+13x^3-32x^2-45x+18=(x-2)(x+3)(2x+3)(3x-1)。

Answer 3

(x+3)(x-2)(2x+3)(3x-1)