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What does it mean by the term "x-polynomial"?

Let say I have a problem like this:

Consider the polynomial $2x^2-5xy-3y^2+x+11y-6$ as $x – polynomial$. Find the degree and numerical coefficient of each term.

In my understanding, this means that all “y” variables are considered constant in the given polynomial. Thus, we can transform the polynomial to $2x^2+(-5y+1)x+(-3y^2+11y-6)$.

AYA
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    Just right. You think of the y’s as constants for this purpose. – Lubin Apr 09 '20 at 02:51
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    What you stated is my understanding as well. With quite a few phrases, it might depend to some extent on the context where it's being used, but I don't think that potential issue applies to this phrase. – John Omielan Apr 09 '20 at 02:51
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    Your understanding is correct. This is also called a polynomial in $x$. A polynomial in $y$ would be $-3y^2+(11-5x)y+(2x^2+x-6)$. – Andrew Chin Apr 09 '20 at 02:52

2 Answers2

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As stated in the comments, you are correct in saying that all instances of $y$ are treated as constants. The degree of the function is then given by the highest power of $x$ when you rearrange the polynomial as you have done. Therefore, the equation has degree $2$.

Micah Windsor
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What you are doing is the correct way to express it as "x–polynomial".

You arrange it just like a standard polynomial, with $x$ going from highest degree to lowest degree.

You can also factor this expression further...

$2x^2+(-5y+1)x+(-3y^2+11y-6)$

$2x^2+(-5y+1)x-(3y^2-11y+6)$

$2x^2+(-5y+1)x-(3y-2)(y-3)$

$[2x+(y-3)][x-(3y-2)]$

$(2x+y-3)(x-3y+2)$ -----> Factored as far as we can go.

Kepping the $x$ as the first term in each bracket is important for maintaining the whole "x–polynomial" bit.

If $y$ was the variable we needed, we would do the same thing, but using $y$ from highest degree to lowest degree.

TheMathMan
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