The remainder of f(x) divided by (x- 1) is 2, so g (x) = f (x)-2 = px3-11x 2+qx-6-2 = px3-1668.

Then g(x) can be divisible by (x- 1), so (x- 1) is a factor of g(x), that is, x- 1=0 is one of the solutions of the equation g(x)=0, so there is g (65438+.

Similarly, let h (x) = f (x)-14 = px3-1x 2+qx-20, and get 64p- 176+4q-20=0, that is,/kloc-.

Solve equation ① ② about p and q to get p=2 and q= 17.

(2)F(x)= 2x 3- 1 1x 2+ 17x-6 from( 1)。

Factorization: f(x)

=2x^3-x^2- 10x^2+5x+ 12x-6

=x^2(2x- 1)-5x(2x- 1)+6(2x- 1)

=(x^2-5x+6)(2x- 1)

=(x-2)(x-3)(2x- 1)

(3) According to (2), the equation f(x)=0 can be changed into (x-2)(x-3)(2x- 1)=0, so x-2=0, x-3=0, 2x- 1=0, that is.

That's it for today!