Question

f(X)=x^3-x^2-4x+4...Factor this polynomial completely over the set of complex numbers

Answer 1

always check first if \(f(1)=0\)

Answer 2

how do i do that?

Answer 3

replace \(x\) by \(1\)

Answer 4

do it in your head
\[f(1)=1-1-4+4=0\]

Answer 5

yes its equal to zero

Answer 6

once you see that \(f(1)=0\) you can factor as
\[x^3-x^2-4x+4.=(x-1)(something)\]

Answer 7

you can find the "something" by division or synthetic division or thinking
the "something" will be a quadratic, and you can find the zero of that using the quadratic formula if you have to

Answer 8

ok i'll try the quadratic formula so....is a=-1 b=-4 c=4 ?

Answer 9

did you factor yet?

Answer 10

you have to factor before you can use the quadratic formula

Answer 11

factor what?

Answer 12

\[x^3-x^2-4x+4=(x-1)(something)\] you need the "something"

Answer 13

you have to divide \(x^3-x^2-4x+4\) by \(x-1\) to find it

Answer 14

You could also factor by grouping, if I'm not mistaken.

Answer 15

ok so i used synthetic division and i got (x-1)(2x^2-4x+4), correct?

Answer 16

where did the 2 come from?

Answer 17

the leading coefficient is 1, there is no 2 in it

Answer 18

yea but dont in synthetic division you multiply -1 by 1 and than on the top i have -1 + -1=-2

Answer 19

1 -1 -4 4
1
_____________
1

Answer 20

leading coefficient is 1

Answer 21

ok i see what you mean i put -1 as the leading coefficient

Answer 22

no you put 2

Answer 23

ok well....now i got 1 0 -4 0

Answer 24

1 -1 -4 4
1
-1 0 -4
____________
1 0 -4 0

Answer 25

so it is
\[(x-1)(x^2-4)\]

Answer 26

okay thanks!

Answer 27

yw