A polynomial of degree $13$ is divided by $d(x)$ to give a quotient of degree $7$ and a remainder of $3x^3+4x^2-x+12$. What is $\deg d$?
Let $f(x)$ be the polynomial of degree $13$ and let $q(x)$ be the quotient when $f(x)$ is divided by $d(x)$. Let $r(x) = 3x^3+4x^2-x+12$. Then we have
$$f(x) = d(x)\cdot q(x) + r(x).$$where $\deg q = 7$.

Since $\deg r = 3$, we need to have $\deg(d\cdot q) = \deg f$ which means $\deg d + \deg q = \deg f$.  So $\deg d = 13-7 = \boxed{6}.$
6
Level 2
Intermediate Algebra
Problem: A polynomial of degree $13$ is divided by $d(x)$ to give a quotient of degree $7$ and a remainder of $3x^3+4x^2-x+12$. What is $\deg d$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the degree of the divisor polynomial \( d(x) \), we apply the polynomial division theorem. According to this theorem, when a polynomial \( p(x) \) of degree \( n \) is divided by another polynomial \( d(x) \) of degree \( m \), the degree of the quotient \( q(x) \) is given by \( n - m \), and the degree of the remainder \( r(x) \) must be less than the degree of the divisor \( d(x) \).

In this problem, the polynomial \( p(x) \) has a degree of \( 13 \). The quotient \( q(x) \) has a degree of \( 7 \). Therefore, we can set up the relationship:

\[
\deg p = \deg q + \deg d
\]

Substituting the known values:

\[
13 = 7 + \deg d
\]

To isolate \( \deg d \), we rearrange the equation:

\[
\deg d = 13 - 7
\]

Calculating this gives:

\[
\deg d = 6
\]

Next, we consider the remainder \( r(x) = 3x^3 + 4x^2 - x + 12 \). The degree of the remainder is \( 3 \), which is less than the degree of the divisor \( d(x) \). This condition is satisfied since \( \deg d = 6 \) is greater than \( 3 \).

Thus, the degree of the divisor polynomial \( d(x) \) is confirmed to be \( 6 \).<|END_OF_LATENT|> The final answer is $6$.