x^3 – 3x^2 + 0x +4

————————

x – 2

…………x^2

………..————————–

x – 2…/ x^3 – 3x^2 + 0x + 4

………- (x^3 – 2x^2)

………..—————-

……………….. – x^2 + 0x

…………x^2 – x

………..————————–

x – 2…/ x^3 – 3x^2 + 0x + 4

………- (x^3 – 2x^2)

………..—————-

……………….. – x^2 + 0x

………………..- (-x^2 +2x)

………………….——————

…………………..( – 2x + 4 )

…………………x^2 – x – 2

………..————————–

x – 2…/ x^3 – 3x^2 + 0x + 4

………- (x^3 – 2x^2)

………..—————-

……………….. – x^2 + 0x

………………..- (-x^2 +2x)

……………….——————

………………….( – 2x + 4)

………………… – (-2x + 4)

………………….————

……………………….0

(x – 2)(x^2 – x – 2)

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

check your work by multiplying these factors

to see if the result is x^3 – 3x^2 + 4

Check: (x-2)^2 * (x+1)

= (x^2 -4x + 4)(x +1)

(x^3 – 4x^2 + 4x) + (x^2 -4x + 4)

x^3 -3x^2 +4 this seems to work

Note 1:

The trick to testing what to divide by:

(A) Most commonly used factors in math problems are

(X + 1) or (X – 1), (X + 2) or (X – 2)

So I would try one of these first to see if it divides evenly

(B) Since the constant/last term in the equation is “4”

Then the possible factors to try are factors of 4

(+1, -1, +2, -2, +4, -4)

So after (X + 1) and (X – 1) did not work

then I tried dividing by (X + 2) or (X – 2)

Note 2:

After you find that (X – 2) divides evenly,

Check again to see if the remainder can factor.

In this case, remainder (x^2 – x – 2) factored again into

(x+1)(x-2)

The final answer

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

can also be written as

(x + 1)(x – 2)^2

Note 3:

Always check your work.

It is easier to multiply the factors to get the original answer,

than it is to divide.

The hardest thing is watching the minus signs when

you multiply and subtract during steps of long division.

So checking your work will make sure

you do not have multiplication or sign errors.

http://en.wikipedia.org/wiki/Quadratic_formula#Quadratic_formula