Polynomial Long Division: Additional Practice (Set 2)

Divide with $k$ in tow. First pass: $2x^2(x - 2) = 2x^3 - 4x^2$ leaves $3x^2 + kx - 4$. Second pass: $3x(x - 2) = 3x^2 - 6x$ leaves $(k + 6)x - 4$. Third pass: $(k + 6)(x - 2) = (k + 6)x - 2(k + 6)$ leaves the remainder.

$R = -4 + 2(k + 6) = 2k + 8$

Setting $2k + 8 = 10$ gives $k = 1$. Verify: dividing $2x^3 - x^2 + x - 4$ by $x - 2$ gives quotient $2x^2 + 3x + 7$ and remainder $10$, and $(x - 2)(2x^2 + 3x + 7) + 10 = 2x^3 - x^2 + x - 4$.

First pass: $x^2(x + 5) = x^3 + 5x^2$; subtracting leaves $-x^2 - 7x + 2$.

Second pass: $-x(x + 5) = -x^2 - 5x$; subtracting leaves $-2x + 2$. Third pass: $-2(x + 5) = -2x - 10$; subtracting leaves $12$.

$x^3 + 4x^2 - 7x + 2 = (x + 5)(x^2 - x - 2) + 12$

Check: $(x + 5)(x^2 - x - 2) = x^3 + 4x^2 - 7x - 10$, and adding $12$ restores the dividend. The remainder is $12$.

First pass: $6x^4 \div 3x = 2x^3$, and $2x^3(3x + 2) = 6x^4 + 4x^3$; subtracting leaves $-3x^3 + x^2 + 5x + 1$.

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

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

Check: $(3x + 2)(2x^3 - x^2 + x + 1) = 6x^4 + x^3 + x^2 + 5x + 2$, and subtracting $1$ restores the dividend. The remainder is $-1$.

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

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

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

The quotient is $x^2 - 2x + 2$ (with remainder $-6$). Sanity check the degrees: $3 = 1 + 2$, and the leading coefficients: $3 = 3 \cdot 1$.

Write the dividend as $x^5 + 0x^4 + 0x^3 + 0x^2 + 0x - 1$ and watch the pattern: each pass leaves the next power behind.

$x^4(x - 1) = x^5 - x^4$ leaves $x^4 + \cdots - 1$; then $x^3(x - 1)$ leaves $x^3 + \cdots - 1$; then $x^2$, then $x$, then $1$, whose product $x - 1$ subtracts to $0$.

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

Every quotient coefficient is $1$: each subtraction hands its cancellation down one column. The alternating-sign option is the quotient of $x^5 + 1$ by $x + 1$, a different division.

Write the dividend as $8x^3 + 0x^2 + 0x + 1$.

First pass: $8x^3 \div 2x = 4x^2$, and $4x^2(2x + 1) = 8x^3 + 4x^2$; subtracting leaves $-4x^2 + 0x + 1$. Second pass: $-4x^2 \div 2x = -2x$, and $-2x(2x + 1) = -4x^2 - 2x$; subtracting leaves $2x + 1$. Third pass: $1 \cdot (2x + 1) = 2x + 1$; subtracting leaves $0$.

$8x^3 + 1 = (2x + 1)(4x^2 - 2x + 1)$

The division is exact, and it has produced the sum-of-cubes factorization mechanically.

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

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

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

The quotient is $x^2 - 2x + 2$ with remainder $0$: a sum that looks unfactorable splits cleanly once the right quadratic divisor is offered.

The first division says $P = DQ$. Read the same product the other way around:

$P = Q \cdot D + 0$

The remainder $0$ is admissible against any divisor, so this identity satisfies both promises of the division algorithm for division by $Q$, and by uniqueness it is THE answer: quotient $D$, remainder $0$. The two factors of a certified factorization can trade places as divisor and quotient.

Insert the placeholder: $x^3 + 0x^2 - 4x + 6$.

First pass: $x^2(x - 2) = x^3 - 2x^2$; subtracting leaves $2x^2 - 4x + 6$. Second pass: $2x(x - 2) = 2x^2 - 4x$; subtracting leaves $0x + 6 = 6$, and the loop stops at degree $0 < 1$. Notice the quotient's constant term is $0$: the third pass never ran because the $x$ column cancelled completely.

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

Check: $(x - 2)(x^2 + 2x) = x^3 - 4x$, and adding $6$ restores the dividend.

Evaluate the identity at $x = -3$, where the divisor factor vanishes.

$P(-3) = (-3 + 3)\,Q(-3) - 7 = 0 \cdot Q(-3) - 7 = -7$

The unknown $Q(-3)$ is multiplied by zero, so only the constant $-7$ survives.

Write the dividend as $2x^5 - x^4 + 0x^3 + 3x^2 + 0x - 8$.

First pass: $2x^5 \div x^3 = 2x^2$, and $2x^2(x^3 + 2) = 2x^5 + 4x^2$; subtracting leaves $-x^4 - x^2 - 8$. Second pass: $-x^4 \div x^3 = -x$, and $-x(x^3 + 2) = -x^4 - 2x$; subtracting leaves $-x^2 + 2x - 8$, of degree $2 < 3$: stop.

$2x^5 - x^4 + 3x^2 - 8 = (x^3 + 2)(2x^2 - x) + (-x^2 + 2x - 8)$

The quotient is $2x^2 - x$. Check: $(x^3 + 2)(2x^2 - x) = 2x^5 - x^4 + 4x^2 - 2x$, and adding $-x^2 + 2x - 8$ restores the dividend.

Subtract the identities and gather the $D$ terms: $D(Q - Q') = R' - R$.

If $Q \neq Q'$, the left side is a nonzero polynomial of degree at least $\deg D$, because degrees add under multiplication. But the right side, a difference of two polynomials each zero or of degree below $\deg D$, is zero or of degree below $\deg D$. No polynomial can be both, so $Q = Q'$.

$D(Q - Q') = 0 \ \Rightarrow \ R' - R = 0$

Then the remainders agree too. The quotient-remainder pair is unique, which is why we may speak of THE quotient and THE remainder.