What do i do with a polynomial with complex roots?

abel muroi

Junior Member
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Jan 13, 2015
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I was given the problem

y = x4 - 6x3 + 10x2 - 6x + 9

and was told to find all the zeros.. so do you think i should find the zeros with long division?
 
Hello, abel muroi!

I was given the problem: \(\displaystyle \:y \:=\; x^4-6x^3 + 10x^2+6x+9\)
and was told to find all the zeros.
So do you think i should find the zeros with long division? . Yes!

I "eyeballed" the polynomial and saw a way to factor it.

\(\displaystyle x^4 -6x^3 + 10x^2 - 6x+ 9\)

\(\displaystyle \qquad =\; x^4 - 6x^3 + 9x^2 + x^2 - 6x + 9\)

\(\displaystyle \qquad =\; x^2(x^2 - 6x + 9) + x^2-6x+9)\)

\(\displaystyle \qquad =\;x^2(x-3)^2 + (x-3)^2\)

\(\displaystyle \qquad =\;(x-3)^2(x^2+1)\)
 
Hello, abel muroi!


I "eyeballed" the polynomial and saw a way to factor it.

\(\displaystyle x^4 -6x^3 + 10x^2 - 6x+ 9\)

\(\displaystyle \qquad =\; x^4 - 6x^3 + 9x^2 + x^2 - 6x + 9\)

\(\displaystyle \qquad =\; x^2(x^2 - 6x + 9) + x^2-6x+9)\)

\(\displaystyle \qquad =\;x^2(x-3)^2 + (x-3)^2\)

\(\displaystyle \qquad =\;(x-3)^2(x^2+1)\)

ah thank you for your response, so is (x2 + 1) the same thing as (x + 1)2
 
Hello, abel muroi!


I "eyeballed" the polynomial and saw a way to factor it.

\(\displaystyle x^4 -6x^3 + 10x^2 - 6x+ 9\)

\(\displaystyle \qquad =\; x^4 - 6x^3 + 9x^2 + x^2 - 6x + 9\)

\(\displaystyle \qquad =\; x^2(x^2 - 6x + 9) + x^2-6x+9)\)

\(\displaystyle \qquad =\;x^2(x-3)^2 + (x-3)^2\)

\(\displaystyle \qquad =\;(x-3)^2(x^2+1)\)
Responses like this just show me that mathematics is beautiful and is a true art. Thanks for making me smile! (and open my mind a little more)
 
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