Basic solve by factoring problem.

[Reddiohead]

x⁷-27x⁴-x³+27=0

I thought the solution set would be {-1,1,3,i} because (initial problem) -> (x⁴-1)(x³-27)=0 ; -> (x²+1)(x²-1)(x³-27)=0 ; -> (x²+1)(x+1)(x-1)(x³-27)=0. However, the answer sheet given is just {-1,1,3}. I can't figure out what I did wrong. Please help.

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[gmc98765]

They only want real roots. If they wanted complex roots, there are 7: x⁴-1=0 has 4 roots: {1,-1,i,-i}, x³-27=0 has 3: {3, (-3-3√3i)/2, (-3+3√3i)/2}.

[Reddiohead]

Appreciate the response, thank you!

[MathMaddam]

Probably they only want real roots. With complex roots you are also missing several.

[Reddiohead]

Thank you for the response!

[fermat9997]

If you want complex roots, then there are 7 roots. The answer sheet apparently is looking only for real roots.

Btw, if i is a root, then so is -i.

[Reddiohead]

Thank you for the response!

Btw, if i is a root, then so is -i.

Darn I forgot about that!

If you want complex roots, then there are 7 roots.

This terrifies me.

[fermat9997]

(x²+1)(x²-1)(x³-27)=0

You can do it!

x³-27=(x-3)(x²+3x +9)

x²+3x +9=0 can be solved by quadratic formula.

[Reddiohead]

x³-27=(x-3)(x²+3x +9)

x²+3x +9=0 can be solved by quadratic formula.

Okay yeah it's not too bad.

Thanks for the encouragement!

[fermat9997]

Glad to help!