Asked by Bad "All-American" Bunny
use fundamental theorem algebra to find roots 4x^-6x^3+2x^7=8x^5+3x^6
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Answered by
GPT-5 mini
AI
I need to confirm the intended equation — your typing is ambiguous. Do you mean
4x − 6x^3 + 2x^7 = 8x^5 + 3x^6 ?
(You wrote "4x^-6x^3+2x^7=8x^5+3x^6", which I read as 4x − 6x^3 + 2x^7 = 8x^5 + 3x^6. If you meant something else, please correct me.)
If that is the equation, here are the first steps (using the Fundamental Theorem of Algebra to know how many roots to expect):
1. Move everything to one side:
2x^7 − 3x^6 − 8x^5 − 6x^3 + 4x = 0.
2. Factor out the obvious common factor x:
x(2x^6 − 3x^5 − 8x^4 − 6x^2 + 4) = 0.
So x = 0 is one root. By the Fundamental Theorem of Algebra the remaining factor (degree 6) has six roots (real or complex). Do you want me to attempt to factor that degree‑6 polynomial further (analytically or numerically), or did you mean a different original equation?
4x − 6x^3 + 2x^7 = 8x^5 + 3x^6 ?
(You wrote "4x^-6x^3+2x^7=8x^5+3x^6", which I read as 4x − 6x^3 + 2x^7 = 8x^5 + 3x^6. If you meant something else, please correct me.)
If that is the equation, here are the first steps (using the Fundamental Theorem of Algebra to know how many roots to expect):
1. Move everything to one side:
2x^7 − 3x^6 − 8x^5 − 6x^3 + 4x = 0.
2. Factor out the obvious common factor x:
x(2x^6 − 3x^5 − 8x^4 − 6x^2 + 4) = 0.
So x = 0 is one root. By the Fundamental Theorem of Algebra the remaining factor (degree 6) has six roots (real or complex). Do you want me to attempt to factor that degree‑6 polynomial further (analytically or numerically), or did you mean a different original equation?
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