Remainder, Factor Theorem and Partial Fractions

Remainder, Factor Theorem and Partial Fractions Remainder and Factor Theorem Remainder Theorem (a) If a polynomial 𝑃(π‘₯) is divided by a linear divisor π‘₯ βˆ’ 𝑐, the
remainder is 𝑃(𝑐).
(b) If a polynomial 𝑃(π‘₯) is divided by a linear divisor π‘Žπ‘₯ + 𝑏, the
!
!
2 βˆ’
! !
!
when 2π‘₯ ! + 4π‘₯ βˆ’ 1 is
+4 βˆ’
!
!
βˆ’ 1 = βˆ’2
!
!
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
π‘Žπ‘₯ + 𝑏 is a linear factor of the polynomial 𝑃(π‘₯) if and only if
𝑃 βˆ’
Example:
remainder is 𝑃(βˆ’ ).
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 β€œsolve”
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‘₯
=
βˆ’ 2 (π‘₯ + 2)(π‘₯ βˆ’ 2)
π‘₯!
4π‘₯ βˆ’ 1
4π‘₯ βˆ’ 1
4π‘₯ βˆ’ 1
=
=
!
!
βˆ’ 2π‘₯ + π‘₯ π‘₯(π‘₯ βˆ’ 2π‘₯ + 1) π‘₯ π‘₯ βˆ’ 1 !
Step 3: Express in partial fractions, according to the table below.
π‘₯!
Case
Denominator
Contains
Algebraic Fraction
Partial Fractions
1
Distinct linear
factors
𝑝π‘₯ + π‘ž
(π‘Žπ‘₯ + 𝑏)(𝑐π‘₯ + 𝑑)
𝐴
𝐡
+
π‘Žπ‘₯ + 𝑏 𝑐π‘₯ + 𝑑
2
Repeated linear
factors
𝑝π‘₯ + π‘ž
(π‘Žπ‘₯ + 𝑏)!
𝐴
𝐡
+
π‘Žπ‘₯ + 𝑏 (π‘Žπ‘₯ + 𝑏)!
3
Quadratic factor
that cannot be
factorised
𝑝π‘₯ + π‘ž
(π‘Žπ‘₯ + 𝑏)(π‘₯ ! + 𝑐 ! )
𝐴
𝐡π‘₯ + 𝐢
+
π‘Žπ‘₯ + 𝑏 π‘₯ ! + 𝑐 !
Step 3: Equate the original equation to be 0 and solve the
equation accordingly
Special Algebraic Identities
Sum of Cubes: π‘Ž ! + 𝑏 ! = (π‘Ž + 𝑏)(π‘Ž ! βˆ’ π‘Žπ‘ + 𝑏 ! )
Difference of Cubes: π‘Ž ! βˆ’ 𝑏 ! = (π‘Ž βˆ’ 𝑏)(π‘Ž ! + π‘Žπ‘ + 𝑏 ! )
263 Tanjong Katong Rd #01-­β€07, Tel: 67020118 2π‘₯ ! + 2
β†’ πΌπ‘šπ‘π‘Ÿπ‘œπ‘π‘’π‘Ÿ (π‘‘π‘’π‘”π‘Ÿπ‘’π‘’π‘  π‘Žπ‘Ÿπ‘’ π‘‘β„Žπ‘’ π‘ π‘Žπ‘šπ‘’)
π‘₯ ! βˆ’ 5π‘₯ ! + 2π‘₯ βˆ’ 13
!
2π‘₯ + 2
β†’ πΌπ‘šπ‘π‘Ÿπ‘œπ‘π‘’π‘Ÿ (π‘‘π‘’π‘›π‘œπ‘šπ‘–π‘›π‘Žπ‘‘π‘œπ‘Ÿ β„Žπ‘Žπ‘  β„Žπ‘–π‘”β„Žπ‘’π‘Ÿ π‘‘π‘’π‘”π‘Ÿπ‘’π‘’)
!
π‘₯ + 2π‘₯ + 13
!
2π‘₯ + 2
β†’ π‘ƒπ‘Ÿπ‘œπ‘π‘’π‘Ÿ (π‘›π‘’π‘šπ‘’π‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ β„Žπ‘Žπ‘  π‘™π‘œπ‘€π‘’π‘Ÿ π‘‘π‘’π‘”π‘Ÿπ‘’π‘’)
!
4π‘₯ βˆ’ 6π‘₯ + 16
Step 4: Solve the partial fraction by either cover up rule or comparing coefficients or any other
method.