Question

for the polynomial $-2m^2n^3 + 2m^?n^3 + 7n^2 - 6m^4$ to be a binomial with a degree of 4 after it has been fully simplified, which must be the missing exponent on the $m$ in the second term?
○ 0
○ 1
○ 2
○ 4

Explanation:

Step1: Recall polynomial simplification

A binomial has two terms. To simplify the polynomial \(-2m^{2}n^{3}+2m^{?}n^{3}+7n^{2}-6m^{4}\) into a binomial, we need to combine like terms. Like terms have the same variables with the same exponents. The first and second terms have \(n^{3}\), so they can be like terms if the exponents of \(m\) are equal. Let the missing exponent be \(x\), so we want \(-2m^{2}n^{3}\) and \(2m^{x}n^{3}\) to be like terms, so \(x = 2\) to combine them. Then, let's check the degree. The degree of a term is the sum of exponents of its variables. After combining \(-2m^{2}n^{3}+2m^{2}n^{3}=0\) (wait, no, wait: if we want the simplified polynomial to be a binomial with degree 4. Wait, maybe I messed up. Wait, the polynomial after simplification should have two terms. Let's re - think.

Wait, the polynomial is \(-2m^{2}n^{3}+2m^{x}n^{3}+7n^{2}-6m^{4}\). We need to combine terms. Let's consider which terms can be combined. The terms with \(n^{3}\): \(-2m^{2}n^{3}\) and \(2m^{x}n^{3}\). The terms \(7n^{2}\) and \(-6m^{4}\) are of different types. For the polynomial to be a binomial, we need two of the four terms to combine (so that we have two non - zero terms left). Let's assume that the \(7n^{2}\) term is eliminated? No, \(7n^{2}\) has only \(n\), and the other terms have \(m\) and \(n\) or just \(m\). Wait, the degree of the polynomial after simplification should be 4. The degree of a term \(a m^{p}n^{q}\) is \(p + q\), and the degree of a term \(a m^{p}\) is \(p\), degree of \(a n^{q}\) is \(q\).

If we set the exponent of \(m\) in the second term to 2, then \(-2m^{2}n^{3}+2m^{2}n^{3}=0\), which is not helpful. Wait, maybe we want to combine the first and second terms to eliminate one of the other terms. Wait, no. Wait, let's check the degree. The term \(-6m^{4}\) has degree 4 (since the exponent of \(m\) is 4, and there is no \(n\), so degree is 4). The term \(7n^{2}\) has degree 2. The term \(-2m^{2}n^{3}\) has degree \(2 + 3=5\). The term \(2m^{x}n^{3}\) has degree \(x + 3\).

We want the simplified polynomial to be a binomial with degree 4. So, we need to eliminate the term with degree 5. To eliminate the term \(-2m^{2}n^{3}\) (degree 5), we need to combine it with \(2m^{x}n^{3}\) so that their sum is zero? No, that would eliminate both. Wait, no. Wait, if we set \(x = 2\), then \(-2m^{2}n^{3}+2m^{2}n^{3}=0\). Then the polynomial becomes \(7n^{2}-6m^{4}\), which is a binomial. The degree of \(7n^{2}\) is 2, the degree of \(-6m^{4}\) is 4. The degree of the polynomial is the highest degree of its terms, which is 4. And it is a binomial (two terms: \(7n^{2}\) and \(-6m^{4}\))? Wait, but earlier we thought the \(n^{3}\) terms would combine, but if \(x = 2\), they cancel. Let's verify:

If \(x = 2\), the polynomial is \(-2m^{2}n^{3}+2m^{2}n^{3}+7n^{2}-6m^{4}\). Combining the first two terms: \((-2 + 2)m^{2}n^{3}+7n^{2}-6m^{4}=0+7n^{2}-6m^{4}=7n^{2}-6m^{4}\). This is a binomial (two terms: \(7n^{2}\) and \(-6m^{4}\)). The degree of the polynomial is the highest degree of its terms. The degree of \(7n^{2}\) is 2, the degree of \(-6m^{4}\) is 4. So the degree of the polynomial is 4, which matches the requirement.

If \(x = 0\): The second term is \(2m^{0}n^{3}=2n^{3}\). Then the polynomial is \(-2m^{2}n^{3}+2n^{3}+7n^{2}-6m^{4}\). We have four terms, not a binomial.

If \(x = 1\): The second term is \(2m^{1}n^{3}\). The polynomial is \(-2m^{2}n^{3}+2mn^{3}+7n^{2}-6m^{4}\). Four terms, not a binomial.

If \(x = 4\): The second term is \(2m^{4}n^{3}\). The polynomial is \(-2m^{2}n^{3}+2m^{4}n^{3}+7n^{2}-6m^{4}\). Four terms, not a binomial.

So when \(x = 2\), we can combine \(-2m^{2}n^{3}\) and \(2m^{2}n^{3}\) to get 0, leaving us with \(7n^{2}-6m^{4}\), which is a binomial with degree 4.

Step2: Verify the result

After setting the missing exponent to 2, the polynomial becomes \(-2m^{2}n^{3}+2m^{2}n^{3}+7n^{2}-6m^{4}\). Combining the first two terms: \((-2 + 2)m^{2}n^{3}=0\). So the simplified polynomial is \(7n^{2}-6m^{4}\), which is a binomial (two terms). The degree of \(7n^{2}\) is 2, and the degree of \(-6m^{4}\) is 4. The degree of the polynomial is the highest degree among its terms, which is 4. This satisfies the condition of being a binomial with degree 4.

Answer:

2