What is the quotient of (3x⁴ − 4x² + 8x − 1) ÷ (x − 2)?

Step-by-step solution

1. 1

   Write the dividend with every power present. Long division lines terms up by degree, so a missing power has to be held open with a zero coefficient. There is no $x^3$ term, so write

   $3x^4 + 0x^3 - 4x^2 + 8x - 1$

   Skipping this is the single most common source of a wrong quotient here.

2. 2

   Divide the leading terms to get the first quotient term. Each round of long division only ever looks at the two leading terms:

   $\frac{3x^4}{x} = 3x^3$

   Multiply back and subtract:

   $3x^3(x-2) = 3x^4 - 6x^3$

   $(3x^4 + 0x^3 - 4x^2 + \ldots) - (3x^4 - 6x^3) = 6x^3 - 4x^2 + 8x - 1$

3. 3

   Repeat on the new leading term. Now the remainder-so-far leads with $6x^3$:

   $\frac{6x^3}{x} = 6x^2, \qquad 6x^2(x-2) = 6x^3 - 12x^2$

   Subtracting leaves

   $8x^2 + 8x - 1$

4. 4

   Two more rounds finish the quotient. From $8x^2$:

   $\frac{8x^2}{x} = 8x, \qquad 8x(x-2) = 8x^2 - 16x \ \Rightarrow\ 24x - 1$

   and from $24x$:

   $\frac{24x}{x} = 24, \qquad 24(x-2) = 24x - 48 \ \Rightarrow\ 47$

   The leftover $47$ has degree $0$, lower than the divisor, so the division stops.

5. 5

   Check with the division identity. A correct division must satisfy dividend $=$ divisor $\times$ quotient $+$ remainder:

   $(x-2)(3x^3 + 6x^2 + 8x + 24) + 47 = 3x^4 - 4x^2 + 8x - 1$

   Expanding the left side reproduces the dividend term for term. Equivalently, the remainder theorem gives $f(2) = 48 - 16 + 16 - 1 = 47$, matching the remainder exactly.