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

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

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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#