3條因式定理(急問)

1.
f(x)=x^3-8x^2+17x-10 ---(1)
f(1)=1-8+17-10=0
Let f(x)=(x-1)(ax^2 +bx +c)=ax^3 +(b-a)x^2 +(c-b)x -c ---(2)
By comparing eqt. (1) & (2), we found that
a=1,b=-7,c=10
So,
f(x)=(x-1)(x^2 -7x +10)=(x-1)(x-2)(x-5)

2a.
P(x)=x^3+4x^2-x+k
P(-2)=0
So,
-8 +16 +2 +k=0 and k= -10

2b.
P(x)=x^3+4x^2-x-10 ---(1)
Let P(x)=(x+2)(ax^2 +bx +c)=ax^3 +(b+2a)x^2 +(c+2b)x +2c ---(2)
By comparing eqt. (1) & (2), we found that
a=1,b=2,c=(-5)
So,
P(x)=(x+2)(x^2 +2x -5)
To simply x^2 +2x -5 =0
x= [-2 +/- sqrt.(4+20)]/2 = (-2 +/- sqrt.24)/2 = -1 +/- sqrt.6
x= (-1 +sqrt.6) or (-1 -sqrt.6)
Therefore,
P(x)=(x+2)(x+1 - sqrt.6)(x+1 + sqrt.6)

3a.
f(x)=6x^3-17x^2+x+10
f(5/2)= 6(125/8) - 17(25/4) +5/2 +10 = 375/4 - 425/4 +10/4 +40/4 =0
Therefore,
f(x) is divisible by (2x-5)

3b.
f(x)=6x^3-17x^2+x+10 ---(1)
Let f(x)=(2x-5)(ax^2 +bx +c)=2ax^3 +(2b-5a)x^2 +(2c-5b)x -5c ---(2)
By comparing eqt. (1) & (2), we found that
a=3,b=(-1),c=(-2)
So,
f(x)=(2x-5)(3x^2 -x -2)=(2x-5)(3x+2)(x-1)