What Would You Do with This Multinomial?

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The discussion revolves around the confusion surrounding the expression 5b^2y + x^2y^2 + 5cz^2, which lacks clear instructions for "solving." Participants agree that without an equation or inequality, the term "solve" is misapplied, as one cannot solve an expression. The clarification that the numbers are exponents does not change the fact that the problem is poorly posed. The consensus is that factoring is limited due to the few common factors present in the terms. Overall, the expression requires more context to determine the appropriate mathematical action.
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If someone gave you this 5b2y + x2y2 + 5cz2 and said "Solve this problem" with no further instructions, what in the world would you do with this? It's some sort of multinomial and factoring that out is over my head if that's what is intended here. Anyone have a clue about this?
 
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If they gave me no additional information I’d ask them to tell me what they mean by “solve” since there isn’t even any type of equality. Are the numbers supposed to be exponents?
 
Yes, my apologies. All the 2s mean to the power of 2. So it should read: 5b^2y + x^2y^2 + 5cz^2
 
If he wants you to factor it, you can’t hardly do anything since there are only a few terms with common factors…
 
This is a very poorly posed problem. You cannot "solve" an expression, which is what 5b2y + x2y2 + 5cz2 is.

You can solve an equation (e.g., 2x2 + 3x + 5 = 0) or an inequality (e.g., 2x + y <= 0), but what you have isn't an equation or an inequality.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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