Polynomial Long Division and Placeholder Notation

Fundamentals of Polynomial Long Division

• Division Algorithm for Polynomials:

  • When dividing a dividend polynomial $P(x)$ by a divisor polynomial $D(x)$ where the degree of $D(x)$ is greater than $0$ and less than or equal to the degree of $P(x)$, there exist unique polynomials $Q(x)$ (the quotient) and $R(x)$ (the remainder) such that:     $P(x) = D(x) \times Q(x) + R(x)$
  • Expressed in rational form, the relation is:     $\frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$
  • The degree of the remainder $R(x)$ must be strictly less than the degree of the divisor $D(x)$.

• Degree Alignment and Placeholder Rule:

  • Both the dividend and the divisor must be written in standard descending order of degree: $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$.
  • If any intermediate power of the variable is missing from the dividend sequence, a placeholder term with a coefficient of $0$ must be inserted (e.g., $0x^2$).
  • Inserting placeholder terms ensures vertical column alignment during subtraction steps, preventing arithmetic errors between unlike terms.

Problem 10: Division with Missing Degree Terms

• Problem Statement:

  • Divide the cubic polynomial by the linear binomial:     $(2x^3 - 14x + 10) \div (x + 3)$

• Long Division Setup and Placeholder Insertion:

  • Divisor: $x + 3$
  • Dividend in original form: $2x^3 - 14x + 10$
  • The original dividend lacks a quadratic ($x^2$) degree term.
  • Rewritten dividend with explicit zero placeholder: $2x^3 + 0x^2 - 14x + 10$
  • Division bracket setup:     x + 3 {\overline{\smash{\big)}\,2x^3 + 0x^2 - 14x + 10\phantom{)}}}}

• Step-by-Step Long Division Execution:

  • First Quotient Term:
  • Divide the leading term of the dividend by the leading term of the divisor:       $\frac{2x^3}{x} = 2x^2$
  • Multiply $2x^2$ by the divisor $(x + 3)$:       $2x^2(x + 3) = 2x^3 + 6x^2$
  • Subtract this product from the dividend:       $(2x^3 + 0x^2) - (2x^3 + 6x^2) = -6x^2$
  • Bring down the next term, $-14x$, producing the new partial dividend:       $-6x^2 - 14x$
  • Second Quotient Term:
  • Divide the current leading term by the divisor's leading term:       $\frac{-6x^2}{x} = -6x$
  • Multiply $-6x$ by the divisor $(x + 3)$:       $-6x(x + 3) = -6x^2 - 18x$
  • Subtract this product from the current partial dividend:       $(-6x^2 - 14x) - (-6x^2 - 18x) = 4x$
  • Bring down the constant term, $+10$, producing the new partial dividend:       $4x + 10$
  • Third Quotient Term:
  • Divide the current leading term by the divisor's leading term:       $\frac{4x}{x} = 4$
  • Multiply $4$ by the divisor $(x + 3)$:       $4(x + 3) = 4x + 12$
  • Subtract this product to determine the final remainder:       $(4x + 10) - (4x + 12) = -2$

• Final Result for Problem 10:

  • Quotient: $Q(x) = 2x^2 - 6x + 4$
  • Remainder: $R(x) = -2$
  • Complete rational expression:     $2x^2 - 6x + 4 - \frac{2}{x + 3}$

Problem 11: Quartic Polynomial Division

• Problem Statement:

  • Divide the fourth-degree polynomial by the linear binomial:     $(p^4 + p^3 - 7p^2 - 32p - 16) \div (p - 4)$

• Long Division Setup:

  • Divisor: $p - 4$
  • Dividend: $p^4 + p^3 - 7p^2 - 32p - 16$
  • Because all descending powers of $p$ ($p^4$, $p^3$, $p^2$, $p^1$, $p^0$) are explicitly present, no zero placeholder terms are required.
  • Division bracket setup:     p - 4 {\overline{\smash{\big)}\,p^4 + p^3 - 7p^2 - 32p - 16\phantom{)}}}}

• Step-by-Step Long Division Execution:

  • First Quotient Term:
  • Divide the leading term of the dividend by the leading term of the divisor:       $\frac{p^4}{p} = p^3$
  • Multiply $p^3$ by $(p - 4)$:       $p^3(p - 4) = p^4 - 4p^3$
  • Subtract from the dividend:       $(p^4 + p^3) - (p^4 - 4p^3) = 5p^3$
  • Bring down $-7p^2$:       $5p^3 - 7p^2$
  • Second Quotient Term:
  • Divide the leading term by the divisor's leading term:       $\frac{5p^3}{p} = 5p^2$
  • Multiply $5p^2$ by $(p - 4)$:       $5p^2(p - 4) = 5p^3 - 20p^2$
  • Subtract from the current partial dividend:       $(5p^3 - 7p^2) - (5p^3 - 20p^2) = 13p^2$
  • Bring down $-32p$:       $13p^2 - 32p$
  • Third Quotient Term:
  • Divide the leading term by the divisor's leading term:       $\frac{13p^2}{p} = 13p$
  • Multiply $13p$ by $(p - 4)$:       $13p(p - 4) = 13p^2 - 52p$
  • Subtract from the current partial dividend:       $(13p^2 - 32p) - (13p^2 - 52p) = 20p$
  • Bring down the final constant term, $-16$:       $20p - 16$
  • Fourth Quotient Term:
  • Divide the leading term by the divisor's leading term:       $\frac{20p}{p} = 20$
  • Multiply $20$ by $(p - 4)$:       $20(p - 4) = 20p - 80$
  • Subtract to find the final remainder:       $(20p - 16) - (20p - 80) = 64$

• Synthetic Division Verification for Problem 11:

  • Root of divisor $p - 4 = 0$ is $c = 4$.
  • Row of coefficients: $1$, $1$, $-7$, $-32$, $-16$.
  • Synthetic steps:
  • Bring down the leading coefficient: $1$
  • Multiply by $4$ and add: $1 + (1 \times 4) = 5$
  • Multiply by $4$ and add: $-7 + (5 \times 4) = 13$
  • Multiply by $4$ and add: $-32 + (13 \times 4) = 20$
  • Multiply by $4$ and add: $-16 + (20 \times 4) = 64$
  • Quotient coefficients: $1$, $5$, $13$, $20$; Remainder: $64$

• Final Result for Problem 11:

  • Quotient: $Q(p) = p^3 + 5p^2 + 13p + 20$
  • Remainder: $R(p) = 64$
  • Expressed in standard rational form:     $p^3 + 5p^2 + 13p + 20 + \frac{64}{p - 4}$