Question

alina fully simplifies this polynomial and then writes it in standard form.
$xy^2 - 2x^2y + 3y^3 - 6x^2y + 4xy^2$
if alina wrote the last term as $3y^3$, which must be the first term of her polynomial in standard form?
$\circ\\ xy^2$
$\circ\\ 5xy^2$
$\circ\\ -8x^2y$
$\circ\\ -2x^2y$

Explanation:

Step1: Combine like terms for \(xy^2\)

The terms with \(xy^2\) are \(xy^2\) and \(4xy^2\). Combining them: \(xy^2 + 4xy^2 = 5xy^2\)

Step2: Combine like terms for \(x^2y\)

The terms with \(x^2y\) are \(-2x^2y\) and \(-6x^2y\). Combining them: \(-2x^2y - 6x^2y = -8x^2y\)

Step3: Identify the polynomial in standard form

The polynomial after combining like terms is \(3y^3 + 5xy^2 - 8x^2y\) (since standard form is from highest degree to lowest, and we know the last term is \(3y^3\), so we order by degree of \(x\) or \(y\); here, the highest degree terms considering both variables, but since the last term is \(3y^3\) (degree 3), then the next terms by degree. Wait, actually, when writing in standard form, we usually order by descending degree of the polynomial. Let's check the degrees:
- \(3y^3\): degree 3
- \(5xy^2\): degree \(1 + 2 = 3\)
- \(-8x^2y\): degree \(2 + 1 = 3\)

Wait, but the problem says the last term is \(3y^3\), so the polynomial is written as [terms with higher or equal degree, then \(3y^3\) at the end? Wait no, standard form is descending order of degree. Wait, maybe the degree is considered by the sum of exponents. Let's re - evaluate:

Wait, the original polynomial after combining is \(3y^3+5xy^2 - 8x^2y\). If the last term is \(3y^3\), that means we are writing the polynomial in the order where the last term is \(3y^3\), so we must have arranged the other terms before it. Wait, no, standard form is usually from highest degree to lowest. But if the last term is \(3y^3\), that implies that the polynomial is written as [first terms] \(+ 3y^3\). Wait, maybe I made a mistake in the order. Let's re - combine the terms correctly:

Original polynomial: \(xy^2-2x^2y + 3y^3-6x^2y + 4xy^2\)

Group like terms:

\((xy^2 + 4xy^2)+(-2x^2y-6x^2y)+3y^3\)

Which is \(5xy^2-8x^2y + 3y^3\)

Now, if the last term is \(3y^3\), then the polynomial is written as \(5xy^2-8x^2y + 3y^3\)? No, that would have \(3y^3\) as the last term. Wait, no, in standard form, we order from highest degree to lowest. Let's check the degrees:

- \(5xy^2\): degree \(1 + 2=3\)
- \(-8x^2y\): degree \(2 + 1 = 3\)
- \(3y^3\): degree 3

So all terms are degree 3. Now, when writing in standard form, we can order them by the exponent of \(x\) (descending). Let's see:

- \(-8x^2y\): \(x\) exponent 2
- \(5xy^2\): \(x\) exponent 1
- \(3y^3\): \(x\) exponent 0

But the problem says the last term is \(3y^3\), so the order is from highest \(x\) exponent to lowest, so the first term would be the term with the highest \(x\) exponent before the last term? Wait, no, let's re - read the problem: "If Alina wrote the last term as \(3y^3\), which must be the first term of her polynomial in standard form?"

So the polynomial is [first term] + [second term] + \(3y^3\). Let's go back to combining like terms:

We had \(xy^2+4xy^2 = 5xy^2\)

\(-2x^2y-6x^2y=-8x^2y\)

So the simplified polynomial is \(5xy^2-8x^2y + 3y^3\) (if we order by \(x\) exponent from highest to lowest, it would be \(-8x^2y+5xy^2 + 3y^3\)). Wait, now I see my mistake. Let's order by the degree of \(x\) (the exponent of \(x\)) in descending order:

- \(-8x^2y\): exponent of \(x\) is 2
- \(5xy^2\): exponent of \(x\) is 1
- \(3y^3\): exponent of \(x\) is 0

So if the last term is \(3y^3\) (exponent of \(x = 0\)), then the previous term is \(5xy^2\) (exponent of \(x = 1\)) and the first term is \(-8x^2y\) (exponent of \(x = 2\))? Wait, no, let's check the degrees again. The degree of a term \(ax^m y^n\) is \(m + n\).

- For \(-8x^2y\): \(m = 2\), \(n = 1\), degree \(3\)
- For \(5xy^2\): \(m = 1\), \(n = 2\), degree \(3\)
- For \(3y^3\): \(m = 0\), \(n = 3\), degree \(3\)

When writing in standard form, we can order the terms with the same degree by the exponent of \(x\) in descending order. So the order would be:

1. \(-8x^2y\) (highest \(x\) exponent among degree 3 terms)
2. \(5xy^2\) (next \(x\) exponent)
3. \(3y^3\) (lowest \(x\) exponent)

But the problem states that the last term is \(3y^3\), so the polynomial is written as [first term] + [second term] + \(3y^3\). So if we follow the order of descending \(x\) exponent, the first term is \(-8x^2y\)? Wait, no, let's re - do the combination:

Original polynomial: \(xy^2-2x^2y + 3y^3-6x^2y + 4xy^2\)

Combine like terms:

- \(xy^2+4xy^2 = 5xy^2\)
- \(-2x^2y-6x^2y=-8x^2y\)

So the polynomial is \(5xy^2-8x^2y + 3y^3\) (if we order by \(y\) exponent? No, \(3y^3\) has \(y\) exponent 3, \(5xy^2\) has \(y\) exponent 2, \(-8x^2y\) has \(y\) exponent 1. Wait, that's ascending order of \(y\) exponent, which is not standard form. Standard form is descending order of degree (sum of exponents) and then by descending exponent of a variable (usually \(x\)) for same - degree terms.

Since all terms have degree 3, we order by descending exponent of \(x\):

- \(-8x^2y\) (x exponent 2)
- \(5xy^2\) (x exponent 1)
- \(3y^3\) (x exponent 0)

So if the last term is \(3y^3\), the polynomial is \(-8x^2y + 5xy^2+3y^3\)? But that would have \(3y^3\) as the last term. Wait, the problem says "writes it in standard form" and "the last term as \(3y^3\)". So standard form is usually from highest degree to lowest, and for terms with the same degree, we can order by the exponent of \(x\) or \(y\). Let's check the options. The options are \(xy^2\), \(5xy^2\), \(-8x^2y\), \(-2x^2y\).

Wait, maybe I made a mistake in the order. Let's re - express the polynomial:

After combining like terms, we have:

\(3y^3+5xy^2 - 8x^2y\)

Now, if we write this in standard form, and the last term is \(3y^3\), that means we are writing it as [terms] \(+ 3y^3\). So the terms before \(3y^3\) are \(5xy^2\) and \(-8x^2y\). Now, we need to order these two terms. The degree of both is 3. To write in standard form, we can order by the exponent of \(x\) in descending order. So \(-8x^2y\) (x exponent 2) comes before \(5xy^2\) (x exponent 1). So the polynomial is \(-8x^2y + 5xy^2+3y^3\). But the last term is \(3y^3\), so the first term is \(-8x^2y\)? Wait, no, the options include \(-8x^2y\) as an option. Wait, but let's check the degrees again. Wait, maybe the problem is considering the degree of the polynomial as the highest power of \(x\) or \(y\) separately. Wait, no, the degree of a term \(ax^m y^n\) is \(m + n\).

Wait, let's start over:

1. Combine like terms:

- For \(xy^2\) terms: \(xy^2+4xy^2 = 5xy^2\)
- For \(x^2y\) terms: \(-2x^2y-6x^2y=-8x^2y\)

So the polynomial is \(3y^3 + 5xy^2-8x^2y\) (if we order by the exponent of \(y\) in descending order: \(3y^3\) (y exponent 3), \(5xy^2\) (y exponent 2), \(-8x^2y\) (y exponent 1)). But the problem says the last term is \(3y^3\), so the polynomial is written as \(5xy^2-8x^2y + 3y^3\) (order by \(y\) exponent ascending) or \(-8x^2y + 5xy^2+3y^3\) (order by \(x\) exponent descending) or \(3y^3+5xy^2 - 8x^2y\) (order by \(y\) exponent descending).

Wait, the key is that the last term is \(3y^3\), so the polynomial is [first term] + [second term] + \(3y^3\). So we need to find which of the options is the first term. Let's list the terms we have: \(5xy^2\), \(-8x^2y\), and \(3y^3\).

Now, standard form of a polynomial is usually written in descending order of the degree of the polynomial. Since all three terms have degree 3 ( \(3y^3\): \(0 + 3=3\), \(5xy^2\): \(1+2 = 3\), \(-8x^2y\): \(2 + 1=3\) ), we can order them by the exponent of \(x\) in descending order. So:

- \(-8x^2y\) (x exponent 2)
- \(5xy^2\) (x exponent 1)
- \(3y^3\) (x exponent 0)

So if the last term is \(3y^3\) (x exponent 0), the first term is \(-8x^2y\) (x exponent 2).

Answer:

\(\boldsymbol{-8x^2y}\) (corresponding to the option with \(-8x^2y\))