1. **Problem statement:** Find the remainder when dividing the polynomial $4x^3 + 6x^2 + 3x + 4$ by $2x^2 + 1$. 2. **Formula and rules:** When dividing a polynomial $P(x)$ by a polynomial $D(x)$, the remainder $R(x)$ has degree less than the degree of $D(x)$. Here, $\deg(D) = 2$, so $\deg(R) < 2$, meaning $R(x)$ is at most degree 1: $R(x) = ax + b$. 3. **Perform polynomial division:** Divide $4x^3 + 6x^2 + 3x + 4$ by $2x^2 + 1$. - First term: $\frac{4x^3}{2x^2} = 2x$. - Multiply divisor by $2x$: $(2x)(2x^2 + 1) = 4x^3 + 2x$. - Subtract: $(4x^3 + 6x^2 + 3x + 4) - (4x^3 + 2x) = 6x^2 + (3x - 2x) + 4 = 6x^2 + x + 4$. - Next term: $\frac{6x^2}{2x^2} = 3$. - Multiply divisor by $3$: $3(2x^2 + 1) = 6x^2 + 3$. - Subtract: $(6x^2 + x + 4) - (6x^2 + 3) = x + 1$. 4. **Remainder:** The remainder is $x + 1$, which has degree 1, less than degree 2 of divisor. **Final answer:** The remainder is $$x + 1$$.