MHB Help finding multiplicity and zeros?

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The expression (x+4)(x-2)^3(x^2+2x-8) can be factored to reveal its multiplicities and zeros. The zeros are -4 with a multiplicity of 2 and 2 with a multiplicity of 4. The quadratic factor simplifies to (x-2)(x+4), confirming the multiplicities. Thus, the complete factorization is (x+4)^2(x-2)^4. Understanding these factors is essential for analyzing the function's behavior.
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(x+4) (x-2)^3 (x^2+2x-8)

would it be -4 multiplicity of 2
and 2 multiplicity of 4?
 
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Yes, that's correct since the quadratic factor is the product of the other two linear factors.
 
in other words,
(x+ 4)(x- 2)^3(x^2- 2x- 8)= (x+ 4)(x- 2)^3(x- 2)(x+ 4)= (x+ 4)^2(x- 2)^4
as I am sure you realized.
 
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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