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.
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?
3 answers
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?
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
Check out web resources such as:
http://www.purplemath.com/modules/synthdiv2.htm
or
http://www.youtube.com/watch?v=bZoMz1Cy1T4
1 answer