Using the given zero, find all the zeroes and write a linear factorization of f(x)
1 + i is a zero of f(x)= x4-2x^3-x^2 + 6x -6
I did synthetic division and I got that it wasn't a zero?
This is a trick question.
We know that complex roots always come in pairs. If 1+i is a root, so is 1-i, its conjugate.
The given root and its conjugate multiplied together become a real factor, namely (x-1-i)(x-1+i)=x^2-2*x+2.
Repeat your synthetic division using the combined real factor and you will have a quotient of x²-3.
oh yeah, I forgot about the conjugates! thanks!
One more question, how do i divide x^4-2x^3-x^2+6x-6 by x^2-2x+2 using synthetic division?
It's a similar process as long division with numbers.
Check out web resources such as:
http://www.purplemath.com/modules/synthdiv2.htm
or
http://www.youtube.com/watch?v=bZoMz1Cy1T4
To find the zeros of a polynomial, you can use the fact that if a number is a zero of a polynomial, then the polynomial will be equal to zero when that number is substituted for the variable. In this case, you are given that 1 + i is a zero of the polynomial f(x) = x^4 - 2x^3 - x^2 + 6x - 6.
To check if the given value 1 + i is a zero, substitute it into the polynomial equation:
f(1 + i) = (1 + i)^4 - 2(1 + i)^3 - (1 + i)^2 + 6(1 + i) - 6
Now, simplify this expression using the properties of complex numbers:
(1 + i)^4 = (1 + i)(1 + i)(1 + i)(1 + i)
= (1 + 2i + i^2)(1 + 2i + i^2)(1 + 2i + i^2)
= (1 + 2i - 1)(1 + 2i - 1)(1 + 2i + i^2)
= 4i(1 + 2i + i^2)
= 4i(1 + 2i - 1)
= 4i(2i)
= 8i^2
= 8(-1)
= -8
(1 + i)^3 = (1 + i)(1 + i)(1 + i)
= (1 + 2i + i^2)(1 + i)
= (1 + 2i - 1)(1 + i)
= 4i(1 + i)
= 4i + 4i^2
= 4i + 4(-1)
= 4i - 4
= -4 + 4i
(1 + i)^2 = (1 + i)(1 + i)
= (1 + 2i + i^2)
= (1 + 2i - 1)
= 2i
Substituting these simplifications back into f(1 + i):
f(1 + i) = -8 - 2(-4 + 4i) - (2i) + 6(1 + i) - 6
= -8 + 8 - 8i - 2i + 6 + 6i - 6
= 0
Since f(1 + i) is equal to zero, we can conclude that 1 + i is indeed a zero of the polynomial f(x).
Now that we have confirmed that 1 + i is a zero, we can use polynomial division (synthetic division) to find the other zeroes and obtain the linear factorization of f(x). Perform division by dividing f(x) by (x - (1 + i)):
x^3 + (-1 - 2i)x^2 + (1 + 2i)x - (1 + i)
_______________________________________________
x - (1 + i) | x^4 - 2x^3 - x^2 + 6x - 6
Apply synthetic division:
x^3 + (-1 - 2i)x^2 + (1 + 2i)x - (1 + i)
---------------------------------------------------
x - (1 + i) | x^4 - 2x^3 - x^2 + 6x - 6
- (x^4 - (1 + i)x^3)
------------------------
(2i - 1)x^3 + x^2 + 6x - 6
- ((2i - 1)x^3 - (2i - 1)(1 + i)x^2)
---------------------------------------
2ix^2 + 6x - 6
- (2ix^2 - 2ix^2)
-----------------------
8x - 6
- (8x - 8)
---------
2
The result of this division is 2, so there is no remainder.
The quotient obtained from synthetic division is x^3 + (-1 - 2i)x^2 + (1 + 2i)x - 2. This represents the linear factorization of f(x) using the zero 1 + i:
f(x) = (x - (1 + i))(x^3 + (-1 - 2i)x^2 + (1 + 2i)x - 2)
Therefore, the linear factorization of f(x) is:
f(x) = (x - (1 + i))(x^3 + (-1 - 2i)x^2 + (1 + 2i)x - 2)