Factor #x^4 - 2x^3 + 4x^2 - 6x + 3# into linear and quadratic terms knowing that #-isqrt(3)# is a zero?

Question

I think synthetic division might help.
And also this is the real question, ignore the other question that looks very similar.
Thanks.

1638 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-09T23:21:14

#(x+isqrt3)(x-isqrt3)(x-1)^2#

If #x=-isqrt3# is zero of this polynomial, #x=isqrt3# also zero of it. Hence #(x^2+3)# is multiplier of it.

#x^4-2x^3+4x^2-6x+3#

=#x^4+3x^2-2x^3-6x+x^2+3#

=#x^2*(x^2+3)-2x*(x^2+3)+x^2+3#

=#(x^2+3)*(x^2-2x+1)#

=#(x+isqrt3)(x-isqrt3)(x-1)^2#