Jφss Sticks Tuition Centre

Sergeant Loi’s Mid-Year Boot Camp 2008 – Defeat Partial Fractions In Three Devastating Moves

Tuition given in the topic of A-Maths Tuition Notes & Tips from the desk of Sergeant Loi at 1:33 am (Singapore time)

Updated on 13 May 2008

Within its pages lies the secret
to defeating Partial Fractions

*Puts on helmet*

*SHOWS STERN & MEAN FACE*

*Sounds bugle* EVERYBODY FALL IN!

Today you’ll learn to deal with a new enemy originally from the A-Level campaign. Something never been seen by your previous O-Level comrades, nor has it appeared in the pages of the old Ten-Year Series 武林秘籍 (ala Secret Manual).

But fret not! Though it looks formidable at first, its size also makes it unwieldy and predictable, so master the rules of engagement (see A. and B. below) and follow the standard 三大招式 (Three Devastating Moves) detailed in each of the Sample Questions below, and you’ll be on your way to defeating these Partial Fractions!

NOW ALL TURN TO PAGE 1 OF YOUR 武林秘籍 (2008 revised edition)!!!

*Cracks whip*

Frag The Partial Fractions!

A. COMPULSORY PRE-CHECK

The degree of a polynomial is the highest power of x. e.g.
p(x) = 4x³ – 12x² – x – 4 ⇒ degree = 3
q(x) = (x – 1)(x + 2) ⇒ degree = 2 ( ∵ when expanded q(x)=x²+x-2)
For :
1. if degree [ p(x) < q(x) ] ⇒ PROPER ⇒ proceed with main Partial Fraction calculations (YAY!)
2. if degree [ p(x) ≥ q(x) ] ⇒ IMPROPER ⇒ Long Division Time! (see Qn 2 below) (SIANZ!) (note it’s greater OR EQUAL)

B. THE RULES OF PARTIAL FRACTIONS ENGAGEMENT

For : (*Have you ‘propered’ this first? See above.)

Rule	If q(`x`) contains	Partial Fraction must contain
1	linear factor (`ax`+`b`)	for each linear factor
2	repeated linear factor (`ax`+`b`)²	$A/{ax+b} + {B/(ax+b)^2}$ for each repeated linear factor
3	quadratic factor `x`²+`c`²	${Ax+B}/{x^2+c^2}$ for each quadratic factor

(x² – b²) can be factorized further into (x + b)(x – b)
→ two linear factors! (≠ quadratic/repeated linear factor!)

SAMPLE PRACTICE QUESTIONS

(Note the standard 三大招式 steps).

Express ${7x+4}/{(2x+1)(x-2)^2}$ in partial fractions.
Ans:
1. 第一式 (绝招)
  Check: degree of (7x+4) = 1 < degree of (2x+1)(x-2)² = 3
  ⇒ PROPER (YAY!)
2. 第二式
  So ${7x+4}/{(2x+1)(x-2)^2}$
  → 1 x linear factor (2x+1) and
  → 1 x repeated linear factor (x-2)²
  Using Rules 1 & 2 from the table above, we have:
  ${7x+4}/{(2x+1)(x-2)^2} = A/{2x+1} + B/{x-2} + C/(x-2)^2$
  *Note there’re two components for the repeated linear factor (x-2)²!
3. 第三式 (绝招)
  Multiply both sides by (2x+1)(x-2)²,
  7x+4 = A(x-2)² + B(2x+1)(x-2) + C(2x+1)
  To find A, B and C, sub in suitable values for x that makes certain components disappear:
  Sub x = 2,
  
  Sub x = ,
  
  Sub x = 0, A = , C = ,
  
  Hence,
  ${7x+4}/{(2x+1)(x-2)^2}$
  ${} = 2/{25(2x+1)} - 1/{25(2x+1)(x-2)} + 18/{5(x-2)^2}$
Express ${x^4+9}/{x^3+3x}$ in partial fractions.
Ans:
1. 第一式
  Check: degree of (x⁴+9) = 4 > degree of (x³+3x) = 3
  ⇒ IMPROPER ⇒ LONG DIVISION TIME!!! (SIANZ! 🙁 )
2. 第二式
  Now ${x^4+9}/{x^3+3x}$ can be factorized further to ${x^4+9}/{x(x^2+3)}$
  → 1 x linear factor (x) and
  → 1 x quadratic factor (x²+3)
  Using Rules 1 & 3 from the table above, we have:
  ${x^4+9}/{x(x^2+3)} = x + A/x + {Bx+C}/{x^2+3}$
  *Note: all we need is the quotient (x) from the long division in i.
3. 第三式
  Multiply both sides by x(x²+3),
  x⁴+9 = x²(x²+3) + A(x²+3) + (Bx+C)x
  To find A, B and C, sub in suitable values for x that makes certain components disappear or your life easier:
  Sub x = 0,
  ⇒ 9 = (0)(3) + A(3) + (C)(0)
  ⇒ A = 3
  Oh we can’t reduce other components to 0 with another substitution → need simultaneous equations here:
  Sub x = 1 (to make your life easier), A = 3,
  ⇒ 1+9 = (1)(4) + (3)(4) + B + C
  ⇒ B + C = -6 —– (1)
  Sub x = -1 (to make your life easier), A = 3,
  ⇒ 1+9 = (1)(4) + (3)(4) + (-B + C)(-1)
  ⇒ B – C = -6 —– (2)
  Solving (1) & (2),
  B = -6, C = 0
  Hence,
  ${x^4+9}/{x^3+3x} = x + 3/x - {6x}/{x^2+3}$

As always, get these rules drilled into your head! Spot the pointers and common mistakes in red! Understand the representative sample questions!

Print this out if necessary and remember the above procedures by heart, for if you don’t Sergeant Loi will unleash her own version of the 三大招式 upon you!

*Cracks whip!*

頑張って!!!

Sergeant Loi