find roots x^4-4x^3+7x^2-4x+6=0

Question

jamesj · Answer

Try +1, -1, +2, -2; and if none of them work, try +i, -i.

jamesj · Answer

Notice btw that if +i is a root, then -i must also be a root because all the coefficients of the polynomial are real. That being the case, you can now factor out a quadratic and the problem becomes easy.

anonymous · Answer

can we use synthetic division with i???

anonymous · Answer

reply please.........

anonymous · Answer

i is the root. i have applied synthetic division and get a very complicated ans
i.e (x-i)(x^3+(-4+i)x^2+(6-4i)x+6i)=0
how can i do further?

jamesj · Answer

If i is a root, then -i is a root and hence
$$ (x+i)(x=-i) = x^2 + 1 $$
is a divisor of the polynomial; i.e., you can divide the polynomial by $ x^2 + 1 $.

jamesj · Answer

$$ (x+1)(x-i) = x^2 + 1 $$

anonymous · Answer

my method is correct?

jamesj · Answer

Probably, but unduly complicated.  Remember, if a polynomial has all real coefficients, and a complex number $ a $ is a root, then the complex conjugate of $ a $ is also a root and hence $ x^2 + |a|^2 $ is always a factor.

jamesj · Answer

If you factor out $ x^2 + 1$ you will find a much simpler polynomial to deal with: just a quadratic.

anonymous · Answer

(x+1)(x−i)=x^2+1 how do you do this step?

anonymous · Answer

please explain if u can?

jamesj · Answer

In general, $ (a-b)(a+b) = a^2 - b^2 $ thus
$$ (x+i)(x-i) = x^2 - i^2 = x^2 -(-1) = x^2 + 1 $$