Divide 2x^2 + 2x + 1 by 2x + 3. Give the quotient and the remainder.

Step-by-step solution

1. 1

   Set up the division. Every power of $x$ is already present in $2x^2+2x+1$, so no placeholder is needed. The divisor $2x+3$ is non-monic, which means fractional quotient coefficients are normal here.

2. 2

   First round. $\dfrac{2x^2}{2x}=x$. Multiply back: $x(2x+3)=2x^2+3x$. Subtract: $(2x^2+2x+1)-(2x^2+3x)=-x+1.$

3. 3

   Second round. $\dfrac{-x}{2x}=-\dfrac12$. Multiply back: $-\tfrac12(2x+3)=-x-\tfrac32$. Subtract: $(-x+1)-\left(-x-\frac32\right)=1+\frac32=\frac52.$

4. 4

   Stop at the right moment. The leftover $\tfrac52$ has degree $0$, below the degree of $2x+3$, so the process ends: quotient $x-\tfrac12$, remainder $\tfrac52$.

5. 5

   Verify. $\left(x-\frac12\right)(2x+3)+\frac52=2x^2+3x-x-\frac32+\frac52=2x^2+2x+1.$ Notice the quotient matches the previous problem while the remainder does not — changing the divisor constant only moves the leftover.