Polynomial Long Division: Practice

First pass: $x^4 \div x^2 = x^2$, and $x^2(x^2 + 2) = x^4 + 2x^2$; subtracting leaves $-2x^3 + x^2 - x + 4$.

Second pass: $-2x^3 \div x^2 = -2x$, and $-2x(x^2 + 2) = -2x^3 - 4x$; subtracting leaves $x^2 + 3x + 4$. Third pass: $1 \cdot (x^2 + 2) = x^2 + 2$; subtracting leaves $3x + 2$, of degree $1 < 2$: stop.

$x^4 - 2x^3 + 3x^2 - x + 4 = (x^2 + 2)(x^2 - 2x + 1) + (3x + 2)$

Check: $(x^2 + 2)(x^2 - 2x + 1) = x^4 - 2x^3 + 3x^2 - 4x + 2$, and adding $3x + 2$ restores the dividend.

Divide, carrying $k$ along. First pass: $x^2(x + 3) = x^3 + 3x^2$ leaves $-x^2 + kx + 6$. Second pass: $-x(x + 3) = -x^2 - 3x$ leaves $(k + 3)x + 6$. Third pass: $(k + 3)(x + 3) = (k + 3)x + 3(k + 3)$ leaves the remainder.

$R = 6 - 3(k + 3) = -3k - 3$

Exact division demands $R = 0$, so $-3k - 3 = 0$ and $k = -1$. Verify: with $k = -1$ the division of $x^3 + 2x^2 - x + 6$ by $x + 3$ gives quotient $x^2 - x + 2$ and remainder $0$, and $(x + 3)(x^2 - x + 2) = x^3 + 2x^2 - x + 6$.

Rebuild with $P = DQ + R$. First the product:

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

Then add the remainder: $(x^4 + x^3 - x^2 - 3x - 6) + (4x - 1) = x^4 + x^3 - x^2 + x - 7$.

The first option is the bare product $DQ$ with the remainder forgotten, a classic slip.

An identity alone is not a division answer: the division algorithm also demands $R = 0$ or $\deg R < \deg D$.

$\deg(x + 3) = 1 = \deg(x + 1), \qquad \deg 2 = 0 < 1$

The second identity's remainder $x + 3$ is not smaller in degree than the divisor, so another pass of the algorithm could still run on it (indeed $x + 3 = (x + 1) + 2$, which is where the first identity comes from). Only the pair $Q = x + 3$, $R = 2$ keeps both promises, and by the uniqueness theorem it is the only pair that does.

Degrees add under multiplication, and the remainder cannot reach the leading term, so $\deg P = \deg D + \deg Q$.

$\deg D = 7 - 4 = 3$

The remainder is zero or of degree strictly below $\deg D = 3$, so a nonzero remainder has degree at most $2$.

The rewriting divides both sides by $D(x)$, and dividing by zero is meaningless, so every zero of $D$ must be excluded.

$\frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}, \qquad D(x) \neq 0$

The identity form $P = DQ + R$ holds for every $x$ with no exclusions, which is why proofs prefer it; the fraction form inherits holes at the divisor's zeros.

Divide, carrying $c$. First pass: $x^2(x^2 + 3) = x^4 + 3x^2$ leaves $x^3 + (c - 3)x^2 + 3x + 6$. Second pass: $x(x^2 + 3) = x^3 + 3x$ leaves $(c - 3)x^2 + 6$: the $x$ terms cancel on their own. Third pass: $(c - 3)(x^2 + 3) = (c - 3)x^2 + 3(c - 3)$ leaves the remainder.

$R = 6 - 3(c - 3) = 15 - 3c$

Exactness needs $15 - 3c = 0$, so $c = 5$. Verify: $(x^2 + 3)(x^2 + x + 2) = x^4 + x^3 + 5x^2 + 3x + 6$.

First pass: $4x^4 \div 2x^2 = 2x^2$, and $2x^2(2x^2 - x + 1) = 4x^4 - 2x^3 + 2x^2$; subtracting leaves $-2x^3 + 3x^2 - 3x + 2$.

Second pass: $-2x^3 \div 2x^2 = -x$, and $-x(2x^2 - x + 1) = -2x^3 + x^2 - x$; subtracting leaves $2x^2 - 2x + 2$. Third pass: $1 \cdot (2x^2 - x + 1)$ leaves $-x + 1$, of degree $1 < 2$: stop.

$4x^4 - 4x^3 + 5x^2 - 3x + 2 = (2x^2 - x + 1)(2x^2 - x + 1) + (-x + 1)$

The quotient happens to equal the divisor, and the remainder is $-x + 1$.

First pass: $x^2(x - 1) = x^3 - x^2$; subtracting leaves $-x^2 + 3x - 4$.

Second pass: $-x(x - 1) = -x^2 + x$; subtracting leaves $2x - 4$. Third pass: $2(x - 1) = 2x - 2$; subtracting leaves $-2$.

$x^3 - 2x^2 + 3x - 4 = (x - 1)(x^2 - x + 2) - 2$

Check: $(x - 1)(x^2 - x + 2) = x^3 - 2x^2 + 3x - 2$, and subtracting $2$ restores the dividend. The remainder is $-2$, not $+2$.

The remainder must be the zero polynomial or have degree strictly below the divisor's degree $2$.

$\deg(x^2 + 1) = 2 \not< 2$

If a working remainder were $x^2 + 1$, the loop would not stop: one more pass, with quotient term $x^2 \div 2x^2 = \tfrac{1}{2}$, would still run. Constants and linear polynomials, and $0$ itself, are all legitimate remainders.

The identity holds at every $x$, so evaluate it at $x = 4$, where the divisor factor vanishes.

$P(4) = (4 - 4)\,Q(4) + 9 = 0 \cdot Q(4) + 9 = 9$

Whatever $Q(4)$ is, it is multiplied by zero. The remainder alone survives the substitution.

The division algorithm caps the remainder strictly below the divisor's degree; the dividend's degree is irrelevant to that cap.

$\deg R < \deg D = 2$

A nonzero remainder therefore has degree $0$ or $1$, so the largest possible degree is $1$.