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

Step-by-step solution

1. 1

   Notice that synthetic division is not available here. Synthetic division is a shortcut only for divisors of the form $x-c$, that is degree $1$ and monic. The divisor $x^2+2x-3$ has degree $2$, so the general algorithm — polynomial long division — is required. Write the dividend with all powers present:

   $x^3+0\cdot x^2+x+1$

2. 2

   Divide the leading terms to get the first quotient term. Compare $x^3$ with $x^2$:

   $\frac{x^3}{x^2}=x$

   Multiply the whole divisor by $x$ and subtract:

   $x^3+0x^2+x+1-\left(x^3+2x^2-3x\right)=-2x^2+4x+1$

3. 3

   Repeat on the new leading term. Now compare $-2x^2$ with $x^2$:

   $\frac{-2x^2}{x^2}=-2$

   Multiply the divisor by $-2$ and subtract:

   $-2x^2+4x+1-\left(-2x^2-4x+6\right)=8x-5$

   Subtracting a negative is where sign errors cluster: $4x-(-4x)=8x$ and $1-6=-5$.

4. 4

   Stop when the degree drops below the divisor. The current leftover $8x-5$ has degree $1$, which is less than the divisor degree $2$, so the algorithm terminates:

   $\text{quotient}=x-2,\qquad \text{remainder}=8x-5$

   Note the remainder is a polynomial, not a number — that is normal whenever the divisor has degree $2$ or more.

5. 5

   Record and verify the result.

   $\frac{x^3+x+1}{x^2+2x-3}=x-2+\frac{8x-5}{x^2+2x-3}$

   Multiplying back, $(x^2+2x-3)(x-2)=x^3-2x^2+2x^2-4x-3x+6=x^3-7x+6$, and $x^3-7x+6+(8x-5)=x^3+x+1\;\checkmark$.

6. 6

   Optional cross-check by factoring the divisor. Since $x^2+2x-3=(x+3)(x-1)$, the remainder $8x-5$ can be tested at the divisor roots. At $x=1$: the dividend gives $1+1+1=3$ and the remainder gives $8-5=3\;\checkmark$. At $x=-3$: the dividend gives $-27-3+1=-29$ and the remainder gives $-24-5=-29\;\checkmark$. Two matching values pin down a degree-$1$ remainder completely.