Finding Zeros...desperate help.

[jcshovey]

I'm having a problem with this problem...

f(x) = x^4-4x^3+11x^2-28x+28

I put it in my calculator and i found that one of the zeros is x=2...no big deal. I did the synthetic division and there is no remainder but my answers say another answer is x=+-i(square root of 7).....where is that coming from??!?!?!?!

 

[Gene]

There are two zeros at x=2. Do the synthetic division again on the new equation. That will bring it down to a quadratic.

 

[jcshovey]

you are an angel gene!! i'm not sure how that works, but i got the answer i was lookin for and that works for me!!! thank you dearly!!!!!!!! can i ask 1 more question.....

x^3-4x^2+5x-2/x^2+2 (Using synthetic or long division)

the answer is

x-4+ (3x+6)/(x^2+2)

how'd they do it??????? i understand synthetic and long division...but i couldnt get this one....the x squared threw me off b/c it doesnt divide evenly and i dont know how to synthetically divide it.....

 

[Gene]

If you do it by long division with a divisor of x²+0x+2 it should work out. It's hard to do it here, but if you have trouble I'll give it a try.
Or you can do synthetic division twice here too, using +&- i*sqrt(2) as divisors. Either way you should get 3x+6 as the remainder.

The first factors into
(x-2)(x-2)(x+i*sqrt(7))(x-i*sqrt(7))
That's why you can/must take out (x-2) twice.

 

[soroban]

Hello, jcshovey!

f(x) = x⁴ - 4x³ + 11x² - 28x + 28
.

With a polynomial of four or more terms, factoring "by grouping" may work . . . or not.

We have: .x⁴ - 4x³ + 11x² - 28x + 28
. . . . . . . . . . . . . . . . . / . \
. . . . . . x⁴ - 4x³ + 4x² + 7x² - 28x + 28

Factor: . x²(x² - 4x + 4) + 7(x² - 4x + 4)

Factor: . (x² - 4x + 4)(x² + 7)

Factor: . (x - 2)²(x² + 7)
.