MHB Find x- and y- Intercepts....3

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The discussion focuses on finding the x- and y-intercepts of the equation y = 8x^3 - 6x - 1. The y-intercept is determined to be (0, -1) by substituting x = 0 into the equation. For the x-intercept, the equation is set to zero, leading to 0 = 8x^3 - 6x - 1, but the user expresses difficulty in finding "nice" roots. The term "nice" roots refers to easily calculable or rational solutions. The conversation highlights the challenge of solving cubic equations for their x-intercepts.
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Find the x- and y-intercepts.

y = 8x^3 - 6x - 1

Let x = 0

y = 8(0)^3 - 6(0) - 1

y = -1

The y-intercept is y = -1 and it takes place at the point
(0, -1).

To find the x-intercept, let y = 0.

0 = 8x^3 - 6x - 1

I am stuck here.
 
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I don't think this one has "nice" roots. :(
 
What do "nice" roots look like?
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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