MHB Simplify a+b+c+d+2(ab+ac+ad+bc+bd+cd)+4(abc+abd+acd+bcd)+8(abcd)

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The expression a+b+c+d+2(ab+ac+ad+bc+bd+cd)+4(abc+abd+acd+bcd)+8(abcd) cannot be simplified significantly due to the lack of common factors. It is clarified that this is an expression, not an equation. An online computer algebra system suggests a form of simplification as (1/2)(2a+1)(2b+1)(2c+1)(2d+1) - (1/2). This representation may help in understanding the structure of the expression better. Ultimately, the original complexity remains largely intact despite this alternative form.
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four variables a,b,c,d
here is the equation: a+b+c+d+2(ab+ac+ad+bc+bd+cd)+4(abc+abd+acd+bcd)+8(abcd)
wondering if this can be simplified to something much smaller
 
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dlecl said:
four variables a,b,c,d
here is the equation: a+b+c+d+2(ab+ac+ad+bc+bd+cd)+4(abc+abd+acd+bcd)+8(abcd)
wondering if this can be simplified to something much smaller

The answer to your question is not really because there is not a common factor in the polynomial.
 
Hey, it's not an equation it's an expression!

Here's what an online CAS thinks.
 
dlecl said:
four variables a,b,c,d
here is the equation: a+b+c+d+2(ab+ac+ad+bc+bd+cd)+4(abc+abd+acd+bcd)+8(abcd)
wondering if this can be simplified to something much smaller
You could write it as $\frac12(2a+1)(2b+1)(2c+1)(2d+1) - \frac12$, if that helps.
 
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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