Question

How do i solve x^4 - 2x^3 + 8x^2 - 32x - 128 = 0 in the real number system?

Answer 1

x^4 - 2x^3 + 8x^2 - 32x - 128 = 0

By synthetic division

4 |.1 -2.. 8 -32 -128
.. |..... 4.. 8. 64. 128
... -----------------------
-2|.1.. 2. 16. 32.... 0
.. |.... -2... 0 -32
... -----------------
.... 1.. 0..16... 0

x^2 + 16 = 0
x^2 = -16
x = ±√ ̅(-16)
= ±4√ ̅-1

This is not a real value.
The equation is

(x - 4)(x + 2)(x^2 + 16) = 0

Answers in real values are

x = -2
x = 4

Answer 2

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