What is the quotient of (x³ − 3x² + 5x − 3) ÷ (x − 1)?

Step-by-step solution

1. 1

   Check first whether the division will be exact. By the factor theorem, $x - 1$ divides the cubic exactly when substituting $x = 1$ gives zero:

   $1 - 3 + 5 - 3 = 0$

   It does, so expect a remainder of $0$ and a clean quadratic quotient. Knowing this in advance tells you a nonzero leftover means an arithmetic slip.

2. 2

   First round: divide, multiply, subtract. Divide leading terms:

   $\frac{x^3}{x} = x^2, \qquad x^2(x-1) = x^3 - x^2$

   $(x^3 - 3x^2 + 5x - 3) - (x^3 - x^2) = -2x^2 + 5x - 3$

3. 3

   Second round. The running dividend now leads with $-2x^2$:

   $\frac{-2x^2}{x} = -2x, \qquad -2x(x-1) = -2x^2 + 2x$

   $(-2x^2 + 5x - 3) - (-2x^2 + 2x) = 3x - 3$

   Note the sign: subtracting $+2x$ leaves $5x - 2x = 3x$.

4. 4

   Third round closes the division. From $3x$:

   $\frac{3x}{x} = 3, \qquad 3(x-1) = 3x - 3$

   $(3x - 3) - (3x - 3) = 0$

   The remainder is $0$, exactly as the factor theorem predicted, so the quotient is $x^2 - 2x + 3$.

5. 5

   Confirm by multiplying back. Expand the factorisation the division produced:

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

   This matches the original cubic. As a bonus, $x^2 - 2x + 3$ has discriminant $4 - 12 = -8 < 0$, so $x = 1$ is the only real root of the cubic.