Remainder, Factor Theorem and Partial Fractions Remainder and Factor Theorem Remainder Theorem (a) If a polynomial π(π₯) is divided by a linear divisor π₯ β π, the remainder is π(π). (b) If a polynomial π(π₯) is divided by a linear divisor ππ₯ + π, the ! ! 2 β ! ! ! when 2π₯ ! + 4π₯ β 1 is +4 β ! ! β 1 = β2 ! ! Step 1: Check if the fraction is improper or proper. Step 2: If improper (highest power (degree) of the numerator is equal to or higher than the highest power of the denominator), use long division. divided by 2π₯ + 1 is ! ! Factor Theorem ππ₯ + π is a linear factor of the polynomial π(π₯) if and only if π β Example: remainder is π(β ). Example: Remainder Partial Fractions General Steps: = 0, i.e the remainder is 0. Solving cubic equations General Steps: Step 1: Find the first factor by trial-and-error or using the βsolveβ function of the calculator. Remember to show the working using factor theorem to prove that the first factor is indeed a factor Step 2: Use either long division or comparing coefficients to factorise the cubic equation. Step 3: Factorise the base as much as possible Example: 4π₯ 4π₯ = β 2 (π₯ + 2)(π₯ β 2) π₯! 4π₯ β 1 4π₯ β 1 4π₯ β 1 = = ! ! β 2π₯ + π₯ π₯(π₯ β 2π₯ + 1) π₯ π₯ β 1 ! Step 3: Express in partial fractions, according to the table below. π₯! Case Denominator Contains Algebraic Fraction Partial Fractions 1 Distinct linear factors ππ₯ + π (ππ₯ + π)(ππ₯ + π) π΄ π΅ + ππ₯ + π ππ₯ + π 2 Repeated linear factors ππ₯ + π (ππ₯ + π)! π΄ π΅ + ππ₯ + π (ππ₯ + π)! 3 Quadratic factor that cannot be factorised ππ₯ + π (ππ₯ + π)(π₯ ! + π ! ) π΄ π΅π₯ + πΆ + ππ₯ + π π₯ ! + π ! Step 3: Equate the original equation to be 0 and solve the equation accordingly Special Algebraic Identities Sum of Cubes: π ! + π ! = (π + π)(π ! β ππ + π ! ) Difference of Cubes: π ! β π ! = (π β π)(π ! + ππ + π ! ) 263 Tanjong Katong Rd #01-β07, Tel: 67020118 2π₯ ! + 2 β πΌπππππππ (πππππππ πππ π‘βπ π πππ) π₯ ! β 5π₯ ! + 2π₯ β 13 ! 2π₯ + 2 β πΌπππππππ (πππππππππ‘ππ βππ βππβππ ππππππ) ! π₯ + 2π₯ + 13 ! 2π₯ + 2 β ππππππ (ππ’πππππ‘ππ βππ πππ€ππ ππππππ) ! 4π₯ β 6π₯ + 16 Step 4: Solve the partial fraction by either cover up rule or comparing coefficients or any other method.
© Copyright 2026 Paperzz