desperate algebra 2 help!?

use the trial and error method to find the first factor

x^3 - x^2 + 2x - 2 = 0

solve for x=1

1-1+2-2= 0

[yippee]

factor is (x-1)

synthetic division

all exponential values are present so you do not need a zero

1,-1,2,-2

bring down the 1

1

multiply by 1

add to -1

0

multiply by 1

add to 2

2

multiply by 1

add to -2

0 (remainder)

collect the coefficients

1,0,2

x^2+2

x=+/-i(sqrt2)

(x-1)(x+/-i(sqrt2))

2)

x^3 + 2x^2 - 8x -16 = 0

solve for x=2

8+8-16-16= not zero

solve for x= -2

[only the odd exponents change]

-8+8+16-16= 0

factor is (x+2)

all exponential values are present

1+2-8-16

bring down the 1

1

multiply by -2

add to 2

0

multiply by -2

add to -8

-8

multiply by -2

add to -6

0(remainder)

collect the coefficients

1,0,-8

x^2-8

x=+/-sqrt8

x=+/-sqrt4*sqrt2

x=+/-2sqrt2

(x+2)(x+/-2sqrt2)

3.) 2x^4 - 5x^3 - 17x^2 + 41x - 21 = 0

solve for x=1

2-5-17+41-21= 0

factor is (x-1)

all exponential values are present

2,-5,-17,41,-21

bring down the 2

2

multiply by 1

add to -5

-3

multiply by 1

add to -17

-20

multiply by 1

add to 41

21

multiply by 1

add to -21

0(remainder)

collect the coefficients

2,-3,-21,20

2x^3-3x^2-21x+20

back to trial and error

[can see where 20+3 might equal 21+2]

solve for x=1

2-3-21+20= not zero

x= -1

-2-3+21+20= not zero

x=2

16-12-42+20= not zero

x= -2

-16-12+42+20= not zero

x=4

128-48-84+20= not zero

x= -4

-128-48+84+20= not zero

x=5

1250-625--425-21= not zero

x= -5

= not zero

does not factor that I can see

(x-1)(2x^3-3x^2-21x+20)