Find the quotient and the remainder when 3x^2 - 5 is divided by x + 1.

Step-by-step solution

1. 1

   Insert the missing term before dividing. The dividend has no $x$ term, and long division needs every column present:

   $3x^{2}-5=3x^{2}+0x-5.$

   Skipping this placeholder is the single most common cause of a wrong remainder.

2. 2

   Divide the leading terms.

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

   so the first term of the quotient is $3x$. Multiply back and subtract:

   $\left(3x^{2}+0x\right)-3x(x+1)=\left(3x^{2}+0x\right)-\left(3x^{2}+3x\right)=-3x.$

3. 3

   Bring down and repeat. The working line is now $-3x-5$. Dividing leading terms again:

   $\frac{-3x}{x}=-3,$

   so the quotient gains $-3$. Multiply back and subtract:

   $(-3x-5)-(-3)(x+1)=(-3x-5)-(-3x-3)=-2.$

4. 4

   Stop when the degree drops below the divisor's. The leftover $-2$ has degree $0$, lower than the degree $1$ of $x+1$, so the division is finished:

   $\text{quotient}=3x-3,\qquad \text{remainder}=-2.$

5. 5

   Check by multiplying back.

   $(x+1)(3x-3)+(-2)=3x^{2}-3x+3x-3-2=3x^{2}-5\ \checkmark$

6. 6

   Cross-check with the remainder theorem. Dividing by $x+1$ means evaluating at $x=-1$:

   $f(-1)=3(-1)^{2}-5=3-5=-2,$

   which matches the remainder found by long division.