Algebraic Long Division

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Transcript Algebraic Long Division

C2: Chapter 1 Algebra and
Functions
Dr J Frost ([email protected])
Last modified: 2nd September 2013
Terminology
11 ÷ 4 = 2 π‘Ÿπ‘’π‘š 3
?
dividend
?
divisor
quotient
?
?
remainder
Normal Long Division
38.
11 4 2 3 . 0 0 0 0
33
93
88
5 0
1. We found how many whole
number of times (i.e. the
quotient) the divisor went into
the dividend.
2. We multiplied the quotient
by the dividend.
3. …in order to find the
remainder.
4. Find we β€˜brought down’ the
next number.
2
6x
x+5
- 2x + 3
6x3 + 28x2 – 7x + 15
3
2
6x + 30x
2
– 2x – 7x
– 2x2 – 10x
3x + 15
3x + 15
0
The Anti-Idiot Test:
You can check your solution by
finding (x+5)(6x2 – 2x + 3)
2
3x
x-1
+ 0 –4
3x3 – 3x2 – 4x + 4
3
2
3x – 3x
0 – 4x + 4
– 4x + 4
0
2
2x
+ 3x – 4
x - 4 2x3 – 5x2 – 16x + 10
3
2
2x – 8x
Find the
2
3x – 16x
remainder.
3x2 – 12x
-4x + 10
Q: Is (x-4) a factor of
-4x
+
16
2x3 – 5x2 – 16x + 10?
-6
Exercises
1a
1i
2a
2i
2b
Exercise 1B
Divide π‘₯ 3 + 6π‘₯ 2 + 8π‘₯ + 3 by π‘₯ + 1
? +3
π‘₯ 2 + 5π‘₯
Divide π‘₯ 3 βˆ’ 8π‘₯ 2 + 13π‘₯ + 10 by π‘₯ βˆ’ 5
? βˆ’2
π‘₯ 2 βˆ’ 3π‘₯
Divide 6π‘₯ 3 + 27π‘₯ 2 + 14π‘₯ + 8 by π‘₯ + 4
6π‘₯ 2 + 3π‘₯
? +2
Divide βˆ’5π‘₯ 3 βˆ’ 27π‘₯ 2 + 23π‘₯ + 30 by π‘₯ + 6
βˆ’5π‘₯ 2 +? 3π‘₯ + 5
Exercise 1C
Find the remainder when π‘₯ 3 + 4π‘₯ 2 βˆ’ 3π‘₯ + 2 is
divided by π‘₯ + 5.
βˆ’8
?
Dividing polynomials with β€˜missing’ terms
Divide x3 – 1 by x – 1
How would we write the division?
2
π΄π‘›π‘ π‘€π‘’π‘Ÿ: π‘₯ + ?π‘₯ + 1
For Olympiad enthusiasts:
In general, the difference of two cubes can be factorised as:
π‘₯ 3 βˆ’ 𝑦 3 = π‘₯ βˆ’ 𝑦 π‘₯ 2 + π‘₯𝑦 + 𝑦 2
Dividing polynomials with β€˜missing’ terms
Divide x4 – 16 by (x+2)
π΄π‘›π‘ π‘€π‘’π‘Ÿ: π‘₯ 3 βˆ’ 2π‘₯ 2 ?+ 4π‘₯ βˆ’ 8
Recap
dividend
quotient
8
= 2+
3
divisor
1
3
remainder
Remainder and Factor Theorem
We’re trying to work out the remainder when
we divide a polynomial 𝑓 π‘₯ by π‘₯ βˆ’ π‘Ž
𝑓 π‘₯
π‘Ÿ
=π‘ž π‘₯ +
π‘₯βˆ’π‘Ž
π‘₯βˆ’π‘Ž
𝑓(π‘₯) = (π‘₯ βˆ’ π‘Ž)π‘ž(π‘₯) + π‘Ÿ
So what does f(a) equal?
What if 𝑓 π‘Ž = 0?
Remainder and Factor Theorem
!
Remainder Theorem
For a polynomial 𝑓(π‘₯), the remainder when
𝑓(π‘₯) is divided by π‘₯ βˆ’ π‘Ž is 𝑓 π‘Ž .
!
Factor Theorem
If 𝑓 π‘Ž = 0, then by above, the remainder is
0. Thus (π‘₯ βˆ’ π‘Ž) is a factor of 𝑓 π‘₯ .
Basic Examples
Remainder when π‘₯ 2 + 1 is divided by π‘₯ βˆ’ 2?
𝑓 2 = ?5
Remainder when π‘₯ 3 βˆ’ π‘₯ is divided by π‘₯ + 1?
𝑓 βˆ’1 =?0
Remainder when π‘₯ 2 + 1 is divided by 2π‘₯ βˆ’ 1?
1
5
𝑓
=?
2
4
Remainder when π‘₯ 2 βˆ’ π‘₯ is divided by 3π‘₯ + 4?
4
28
𝑓 βˆ’ =?
3
9
Examples
Show that (x – 2) is a factor of x3 + x2 – 4x - 4
𝑓 2 = 8 + 4? βˆ’ 8 βˆ’ 4 = 0
Examples
Fully factorise 2x3 + x2 – 18x – 9
= (π‘₯ βˆ’ 3)(π‘₯ +?3)(2π‘₯ + 1)
Tip: If f(x) = 2x3 + x2 – 18x – 9, then try f(-1), f(1), f(2), etc. until
one of these is equal to 0.
Examples
Fully factorise π‘₯ 3 + 6π‘₯ 2 + 5π‘₯ βˆ’ 12
= (π‘₯ βˆ’ 1)(π‘₯ +? 3)(π‘₯ + 4)
Given that (π‘₯ + 1) is a factor of 4π‘₯4 βˆ’ 3π‘₯2 + π‘Ž,
find the value of π‘Ž.
π‘Ž = βˆ’1?
Examples
C2 May 2013 (Retracted)
𝒂 = πŸ”,?
𝒃 = βˆ’πŸ’
(𝒙 βˆ’ 𝟐)(πŸπ’™ βˆ’
? 𝟏)(πŸ‘π’™ + 𝟐)
Examples
Exercise 1D
Q1, 2, 4, 6, 8, 10
Recap
Q10) Given that (π‘₯ βˆ’ 1) and (π‘₯ + 1) are factors of
𝑝π‘₯3 + π‘žπ‘₯2 βˆ’ 3π‘₯ βˆ’ 7 find the value of 𝑝 and π‘ž.
𝑝 = 3,?π‘ž = 7
Recap
Find the remainder when
16x5 – 20x4 + 8 is divided by (2π‘₯ βˆ’ 1)
15
π‘…π‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝑖𝑠 ?
2
Bro tip: think what you could make x in
order to make the factor (2x-1) zero.
Recap
When 8π‘₯4 βˆ’ 4π‘₯3 + π‘Žπ‘₯2 βˆ’ 1 is divided by (2π‘₯ +
1) the remainder is 3. Find the value of π‘Ž.
π‘Ž =? 16
Exercises
Le Exercise 1E:
β€’ Q1f, g, h, i
β€’ 2, 4, 6, 8, 10
Le Exercise 1F
β€’ 4, 5, 8, 10, 15.