Polynomial Long Division: Practice

Match the identity to the division algorithm's template, dividend equals divisor times quotient plus remainder.

$P = D \cdot Q + R \quad\text{with}\quad D = x + 1$

The factor multiplying the divisor is $x + 2$, so that is the quotient, and the leftover $3$ is the remainder. The pair is valid because $\deg 3 = 0$ is below $\deg(x + 1) = 1$, and by uniqueness it is the only valid pair.

The division algorithm requires the remainder to be the zero polynomial or to have degree strictly below the divisor's degree.

$R = 0 \ \text{ or } \ \deg R < 1$

A polynomial of degree less than $1$ has degree $0$, which is a nonzero constant. So the remainder is a constant, and $0$ is allowed; it is not required to be $0$.

The stopping rule is about degree: the loop runs while the working remainder has degree at least the divisor's degree, and halts as soon as it drops below.

$\deg 12 = 0 < 1 = \deg(x - 5)$

No multiple of $x - 5$ can cancel a constant without introducing a new $x$ term, so no further pass can make progress. The sign of the remainder and the number of passes taken are irrelevant.

Rebuild the dividend from the division identity, dividend equals divisor times quotient plus remainder.

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

This multiply-back computation is exactly the check you should run after every division.

Rewrite the dividend with a placeholder, $x^2 + 0x - 9$, then divide. First pass: $x^2 \div x = x$, and $x(x - 3) = x^2 - 3x$; subtracting leaves $3x - 9$.

Second pass: $3x \div x = 3$, and $3(x - 3) = 3x - 9$; subtracting leaves $0$.

$x^2 - 9 = (x - 3)(x + 3)$

The quotient is $x + 3$ with remainder $0$: the division certifies the familiar difference-of-squares factorization.

A polynomial's value moves with $x$, so numerical size cannot measure whether the remainder is smaller than the divisor. The measure that does not move with $x$ is the degree.

$R = 0 \ \text{ or } \ \deg R < \deg D$

The zero polynomial is named separately because it is not assigned a degree. This degree condition is what makes the quotient and remainder unique, just as $0 \leq r < b$ does for integers.

In $P = DQ + R$ the remainder is too low in degree to affect the leading term, so the leading term of $P$ equals the leading term of $DQ$, and degrees add under multiplication.

$\deg Q = \deg P - \deg D = 3 - 1 = 2$

A cubic divided by a linear divisor always gives a quadratic quotient; use this as a sanity check on every answer.

First pass: $x^2 \div x = x$, and $x(x + 5) = x^2 + 5x$; subtracting leaves $3x + 15$.

Second pass: $3x \div x = 3$, and $3(x + 5) = 3x + 15$; subtracting leaves $0$.

$x^2 + 8x + 15 = (x + 5)(x + 3)$

The quotient is $x + 3$ with remainder $0$.

First pass: $x^2 \div x = x$, and $x(x + 1) = x^2 + x$; subtracting leaves $3x + 9$.

Second pass: $3x \div x = 3$, and $3(x + 1) = 3x + 3$; subtracting leaves $6$, of degree $0 < 1$: stop.

$x^2 + 4x + 9 = (x + 1)(x + 3) + 6$

Check: $(x + 1)(x + 3) = x^2 + 4x + 3$, and adding $6$ restores $x^2 + 4x + 9$. The quotient is $x + 3$ and the remainder is $6$.

The quotient term $q$ is chosen precisely so that $q \cdot D$ has the same leading term as the working remainder.

$W - qD \ \text{ has no term of } \ \deg W$

The two copies of the leading term cancel exactly, so the new working remainder has strictly smaller degree. That guaranteed drop is why the loop must eventually stop.

With $R = 0$ the division identity loses its correction term.

$P(x) = D(x)\,Q(x)$

That equation displays $P$ as a product of two polynomials, so $D$ (and $Q$) are factors of $P$. Division with zero remainder is how a factorization is certified.

The dividend's degree $1$ is already below the divisor's degree $2$, so the pair $Q = 0$, $R = x + 1$ satisfies both promises of the division algorithm.

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

By uniqueness this is the only valid answer, just as $3 \div 5$ gives quotient $0$ and remainder $3$. The division is perfectly well defined.