Polynomial Long Division: Additional Practice (Set 1)

Write the dividend as $x^4 + 0x^3 - 3x^2 + 2x - 1$.

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

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

The quotient is $x^2 - x - 1$ with remainder $2x - 2$.

Divide, carrying $k$. First pass: $x^2(x - 2) = x^3 - 2x^2$ leaves $-x^2 + kx + 8$. Second pass: $-x(x - 2) = -x^2 + 2x$ leaves $(k - 2)x + 8$. Third pass: $(k - 2)(x - 2) = (k - 2)x - 2(k - 2)$ leaves the remainder.

$R = 8 + 2(k - 2) = 2k + 4$

Exactness needs $2k + 4 = 0$, so $k = -2$. Verify: dividing $x^3 - 3x^2 - 2x + 8$ by $x - 2$ gives quotient $x^2 - x - 4$ and remainder $0$, and $(x - 2)(x^2 - x - 4) = x^3 - 3x^2 - 2x + 8$.

First pass: $x^2(x^2 + 3x - 2) = x^4 + 3x^3 - 2x^2$; subtracting leaves $2x^3 + 5x^2 - 10x + 4$.

Second pass: $2x(x^2 + 3x - 2) = 2x^3 + 6x^2 - 4x$; subtracting leaves $-x^2 - 6x + 4$. Third pass: $-1 \cdot (x^2 + 3x - 2) = -x^2 - 3x + 2$; subtracting leaves $-3x + 2$, of degree $1 < 2$: stop.

$x^4 + 5x^3 + 3x^2 - 10x + 4 = (x^2 + 3x - 2)(x^2 + 2x - 1) + (-3x + 2)$

The remainder is $-3x + 2$; watch both signs in the final subtraction.

The loop makes progress only by lowering the working remainder's degree, and the only way to remove its leading term $ax^m$ is to subtract a multiple of $D$ whose leading term is exactly $ax^m$.

$q \cdot bx^n = ax^m \ \Rightarrow \ q = \frac{a}{b}x^{m-n}$

Any other single-term choice leaves the leading term standing (or introduces a higher one), and the degree would not drop. Nothing about the choice promises a zero remainder or integer coefficients; it promises only strict progress, which is what makes the algorithm terminate.

Write the dividend as $x^5 + x^4 - 2x^3 + 0x^2 + x - 1$ and the divisor as $x^2 + 0x - 1$.

First pass: $x^3(x^2 - 1) = x^5 - x^3$; subtracting leaves $x^4 - x^3 + 0x^2 + x - 1$. Second pass: $x^2(x^2 - 1) = x^4 - x^2$; subtracting leaves $-x^3 + x^2 + x - 1$. Third pass: $-x(x^2 - 1) = -x^3 + x$; subtracting leaves $x^2 - 1$. Fourth pass: $1 \cdot (x^2 - 1)$; subtracting leaves $0$.

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

The division is exact and the quotient is $x^3 + x^2 - x + 1$.

Expose the claim with the multiply-back check: $(x + 1)(x + 3) + 2 = x^2 + 4x + 5$, not $x^2 + 3x + 5$, so the claim is wrong.

Now divide correctly, subtracting the entire product each pass. First pass: $x(x + 1) = x^2 + x$; subtracting leaves $2x + 5$ (the student's $3x + 5$ kept the $x$ that should have been removed). Second pass: $2(x + 1) = 2x + 2$; subtracting leaves $3$.

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

The correct quotient is $x + 2$ and the correct remainder is $3$.

A valid remainder has degree strictly below the divisor's degree, so the divisor must have degree at least $3$.

$\deg(4x^2 + 1) = 2 \ \Rightarrow \ \deg D \geq 3$

The divisors of degree $3$ and $4$ are all possible, but against $x^2 + 5$ (degree $2$) the algorithm would not have stopped: the pass $4x^2 \div x^2 = 4$ could still run. So $x^2 + 5$ is impossible.

First pass: $2x^4 \div 2x^2 = x^2$, and $x^2(2x^2 + x) = 2x^4 + x^3$; subtracting leaves $-6x^2 + 3x - 7$, still of degree $2$, so the loop continues.

Second pass: $-6x^2 \div 2x^2 = -3$, and $-3(2x^2 + x) = -6x^2 - 3x$; subtracting leaves $6x - 7$, of degree $1 < 2$: stop.

$2x^4 + x^3 - 6x^2 + 3x - 7 = (2x^2 + x)(x^2 - 3) + (6x - 7)$

The quotient is $x^2 - 3$; its $x$ coefficient is $0$ because the degree-$3$ and degree-$1$ columns cancelled on their own.

First pass: $9x^4 \div 3x^2 = 3x^2$, and $3x^2(3x^2 - x + 1) = 9x^4 - 3x^3 + 3x^2$; subtracting leaves $-3x^3 + 4x^2 - 2x + 5$.

Second pass: $-3x^3 \div 3x^2 = -x$, and $-x(3x^2 - x + 1) = -3x^3 + x^2 - x$; subtracting leaves $3x^2 - x + 5$. Third pass: $1 \cdot (3x^2 - x + 1)$; subtracting leaves $4$.

$9x^4 - 6x^3 + 7x^2 - 2x + 5 = (3x^2 - x + 1)(3x^2 - x + 1) + 4$

The quotient equals the divisor, so the dividend is that quadratic squared, plus $4$. The remainder is $4$.

The remainder cannot reach the leading term, so the leading term of $P$ equals the leading term of $D$ times the leading term of $Q$: coefficients multiply and degrees add.

$6x^5 = 3x^2 \cdot (\text{leading term of } Q) \ \Rightarrow \ \text{leading term of } Q = 2x^3$

Divide the coefficients ($6 \div 3 = 2$) and subtract the degrees ($5 - 2 = 3$).

First division: $x^2(x - 2) = x^3 - 2x^2$ leaves $-4x^2 + 12x - 8$; then $-4x(x - 2) = -4x^2 + 8x$ leaves $4x - 8$; then $4(x - 2) = 4x - 8$ leaves $0$. Quotient: $x^2 - 4x + 4$.

Second division: $x(x - 2) = x^2 - 2x$ leaves $-2x + 4$; then $-2(x - 2) = -2x + 4$ leaves $0$. Quotient: $x - 2$.

$x^3 - 6x^2 + 12x - 8 = (x - 2)(x^2 - 4x + 4) = (x - 2)(x - 2)(x - 2)$

Both remainders are $0$, and the final quotient is $x - 2$: the dividend was $(x - 2)^3$ all along.

Write the dividend as $4x^5 + 0x^4 - 5x^3 + 6x^2 + 5x - 1$.

First pass: $4x^5 \div 2x^2 = 2x^3$, and $2x^3(2x^2 + x - 1) = 4x^5 + 2x^4 - 2x^3$; subtracting leaves $-2x^4 - 3x^3 + 6x^2 + 5x - 1$. Second pass: $-x^2(2x^2 + x - 1) = -2x^4 - x^3 + x^2$; subtracting leaves $-2x^3 + 5x^2 + 5x - 1$. Third pass: $-x(2x^2 + x - 1) = -2x^3 - x^2 + x$; subtracting leaves $6x^2 + 4x - 1$. Fourth pass: $3(2x^2 + x - 1) = 6x^2 + 3x - 3$; subtracting leaves $x + 2$, of degree $1 < 2$: stop.

$4x^5 - 5x^3 + 6x^2 + 5x - 1 = (2x^2 + x - 1)(2x^3 - x^2 - x + 3) + (x + 2)$

The remainder is $x + 2$.