Polynomial Long Division: Practice

First pass: $x^3 \div x = x^2$, and $x^2(x + 1) = x^3 + x^2$; subtracting leaves $x^2 - 5x - 6$.

Second pass: $x^2 \div x = x$, and $x(x + 1) = x^2 + x$; subtracting leaves $-6x - 6$. Third pass: $-6x \div x = -6$, and $-6(x + 1) = -6x - 6$; subtracting leaves $0$.

$x^3 + 2x^2 - 5x - 6 = (x + 1)(x^2 + x - 6)$

The quotient is $x^2 + x - 6$ with remainder $0$.

First pass: $x^3 \div x = x^2$, and $x^2(x - 2) = x^3 - 2x^2$; subtracting leaves $-2x^2 + x + 6$.

Second pass: $-2x^2 \div x = -2x$, and $-2x(x - 2) = -2x^2 + 4x$; subtracting leaves $-3x + 6$. Third pass: $-3x \div x = -3$, and $-3(x - 2) = -3x + 6$; subtracting leaves $0$.

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

Check the middle pass's sign carefully: subtracting $+4x$ from $+x$ gives $-3x$.

First pass: $6x^2 \div 2x = 3x$, and $3x(2x - 1) = 6x^2 - 3x$; subtracting leaves $-4x + 3$.

Second pass: $-4x \div 2x = -2$, and $-2(2x - 1) = -4x + 2$; subtracting leaves $1$, of degree $0 < 1$: stop.

$6x^2 - 7x + 3 = (2x - 1)(3x - 2) + 1$

Check: $(2x - 1)(3x - 2) = 6x^2 - 7x + 2$, and adding $1$ restores the dividend.

Write the dividend with placeholders: $x^3 + 0x^2 + 0x - 1$.

First pass: $x^2(x - 1) = x^3 - x^2$; subtracting leaves $x^2 + 0x - 1$. Second pass: $x(x - 1) = x^2 - x$; subtracting leaves $x - 1$. Third pass: $1 \cdot (x - 1) = x - 1$; subtracting leaves $0$.

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

The quotient is $x^2 + x + 1$ with remainder $0$: the division has produced the difference-of-cubes factorization by pure algorithm.

Write the dividend as $x^4 + 0x^3 + 0x^2 + 0x - 16$ and divide by the degree-$2$ divisor.

First pass: $x^4 \div x^2 = x^2$, and $x^2(x^2 - 4) = x^4 - 4x^2$; subtracting leaves $4x^2 - 16$. Second pass: $4x^2 \div x^2 = 4$, and $4(x^2 - 4) = 4x^2 - 16$; subtracting leaves $0$.

$x^4 - 16 = (x^2 - 4)(x^2 + 4)$

The quotient is $x^2 + 4$ with remainder $0$.

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

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

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

Against a degree-$2$ divisor a linear remainder is fully reduced. The remainder is $x + 1$.

Insert the missing power first: $x^3 + 0x^2 - 7x + 6$.

First pass: $x^2(x - 1) = x^3 - x^2$; subtracting leaves $x^2 - 7x + 6$. Second pass: $x(x - 1) = x^2 - x$; subtracting leaves $-6x + 6$. Third pass: $-6(x - 1) = -6x + 6$; subtracting leaves $0$.

$x^3 - 7x + 6 = (x - 1)(x^2 + x - 6)$

Without the $0x^2$ placeholder the second pass would have collided unlike powers in one column.

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

$\deg R < 2 \ \Rightarrow \ R \text{ is linear or constant}$

Only $4x - 3$, of degree $1$, qualifies. The other three have degree $2$ or $3$, so at least one more pass of the algorithm could still run on them.

Degrees add under multiplication, and the remainder is too small to disturb the leading term, so the quotient's degree is the difference.

$\deg Q = 5 - 2 = 3$

The remainder must be zero or of degree strictly below $2$, so if it is nonzero its degree is at most $1$.

First pass: $x^2(x + 2) = x^3 + 2x^2$; subtracting leaves $3x^2 + 7x + 2$.

Second pass: $3x(x + 2) = 3x^2 + 6x$; subtracting leaves $x + 2$. Third pass: $1 \cdot (x + 2) = x + 2$; subtracting leaves $0$.

$x^3 + 5x^2 + 7x + 2 = (x + 2)(x^2 + 3x + 1)$

The quotient is $x^2 + 3x + 1$ with remainder $0$.

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

Second pass: $-4x^2 \div 2x = -2x$, and $-2x(2x + 1) = -4x^2 - 2x$; subtracting leaves $6x + 1$. Third pass: $6x \div 2x = 3$, and $3(2x + 1) = 6x + 3$; subtracting leaves $-2$.

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

A negative remainder is perfectly legal: the condition is about degree, not sign. The remainder is $-2$.

In $P = DQ + R$ the remainder has degree below $\deg D$, so it cannot touch the top: the leading term of $P$ is the leading term of $D$ times the leading term of $Q$.

$2x \cdot 3x^2 = 6x^3$

Coefficients multiply and exponents add, so $P$ opens with $6x^3$.