Pre- Cal QUESTION?

Question

Directions:
Find all the zeros of the function and write the poloynomial as a product of linear factors. ( what are they asking me to do math matically)

h(x)= x^3-3x^2+4x-2

can anyone show me how i would solve this..thankyou!

Answer 1

graph and find there is only one real zero which is 1 so (x-1) is one of the factors.
do synthetic division using factor 1 and the polynomial left will be x^2-2x+2
us quad.formula to get zeros for this which will be 1+/- i which are complex
so in factored form you have
(x-1)(x-(1+i))(x-(1-i))

Answer 2

If you graphed this, you'd be finding the points where the curve crosses the x-axis.
You may want to recall your polynomial factoring skillz from your algebra days....
let x^3-3x^2+4x-2 = 0
try out different values for x until you find one that make it equal 0:
if x =1, then 1-3+4-2=0 (HEY THAT WORKED!)
So, the linear factor that will would give you zero when x=1 is (x-1).
So NOW, use polynomial division to divide x^3-3x^2+4x-2 by x-1
which gets x^2-2x+2. This quadratic, however, won't factor. In fact, take the discriminant (b^2-4ac) and see if it even crosses the x-axis at all ((-2)^2-4*1*2 = -4). The negative number tells us that it does NOT. Positive would've meant yes.
So, your simplified function is h(x) = (x-1)(x^2-2x+2) has a single, reall root at x = 1, while the other two roots are imaginary (I used quadratic formula and factored out the negative square root out as i. Then I simplified).... 1+i and 1-i.

Answer 3

The usual way (look at it). What happens if x=1?
That's right, x-1 is a factor. Now divide the cubic by x-1 and see if the quadratic factors.

Answer 4

Possible Roots: -1,-2,1,2 .. from factors of 1 * -2
(x - 1) is one
Divide:
(x - 1)(x^2 - 2x + 2)
x - 1 = 0 and x^2 - 2x + 2 = 0
x = 1 and x^2 - 2x = -2
x = 1 and (x - 1)^2 = -1
x = 1 and x = 1 +- i