## Wednesday, October 19, 2011

1.  State the remainder theorem:                                                                                       [1]

Either “dividend = divisor × quotient + remainder”

Or f(x) = (x – a)Q(x) + R

2.   Solve for p, q and r: 3x2 + 6x – 4 ≡ p(x –q) 2 + r                                                         [3]
The left side is the expanded form. The right side is the vertex form.
We can convert the left to the right or from the right to the left.  Right to left is probably easier.
right side  = p(x –q) 2 + r
right side  = p(x2 –2qx +q2)  + r
right side  = px2 –2pqx +pq2  + r
left side = 3x2 + 6x – 4
By comparison, 3x2 = px2           --> p = 3
By comparison, 6x = -2pqx = -2(3)qx = -6qx             --> q = -1
By comparison, – 4 = pq2  + r                                        --> r = -7

3.  The polynomial P(x) = x3 + ax2 + bx – 3 has a remainder of 27 when divided by    x – 2, and a remainder of 3 when divided by x + 1.   Calculate the remainder when P(x) is divided by x – 1.                                                                                                  [4]

i) x – 2 = 0 à x = 2
When x = 2, f(x) = 27 (Remainder theorem)
f(2) = 27 = 23 + a(2)2 + b(2) – 3
22 = 4a + 2b  (Note all the coefficients are even)
11 = 2a + 1b   (Equation 1)
ii) x +1 = 0 à x = -1
When x = -1, f(x) = 3 (Remainder theorem)
f(-1) = 3 = (-1)3 + a(-1)2 + b(-1) – 3
7 =   1a – 1b  (Equation 2)
Eqn 1 + Eqn 2: 3a = 18 à a =6
Therefore b = -1
P(x) = x3 + 6x2 – x – 3
4. Is 2x – 3 a factor of  3x3 + 2x2  – 7x + 2?                                                                      [2]
2x – 3 = 0
x = 1.5
3(1.5)3 + 2(1.5)2  – 7(1.5) + 2  0
2x - 3 is not a factor.

5.  Solve the equation 2x3 + 3x2 – 32x + 15 = 0.                                           s09p1.q6  [6]
f(x) = (ax ± b)(1x ± c)(1x ± d)
Obviously a = 2.
b × c × d = 15
Candidates are ±1, ±3, ±5.
f(1)  0
f(-1)  0
f(3) = 54 +27 - 96 +15 = 0
(x – 3) is a factor.
f(5)  0
f(-5) = -250 + 75 + 160 + 15 = 0
(x - -5) = (x + 5) is a factor.
By inspection the third factor must be (2x – 1).
The solutions are x = -5, ½, 3.

### Additional Maths Factor and Remainder Theorem

1.  State the remainder theorem:                                                                                       [1]

2.   Solve for p, q and r: 3x2 + 6x – 4 ≡ p(x –q) 2 + r                                                         [3]

3.  The polynomial P(x) = x3 + ax2 + bx – 3 has a remainder of 27 when divided by    x – 2, and a remainder of 3 when divided by x + 1.   Calculate the remainder when P(x) is divided by x – 1.                                                                                                  [4]

4. Is 2x – 3 a factor of  3x3 + 2x2  – 7x + 2?                                                                      [2]

5.  Solve the equation 2x3 + 3x2 – 32x + 15 = 0.                                                              [6]

## Saturday, October 1, 2011

### Map Ratio

Johnny and Mary read the following question:

“On a map of scale 1 : 20 000 the area of a forest is 50 cm2. On another map the area of the forest is 8 cm2. Find the scale of the second map.”

(Extended Mathematics for IGCSE. David Rayner. Oxford University Press ISBN 978-0-19-913874-6 )

Mary starts with the linear map scale. She knows that 1 : 20 000 is a 'unitless' ratio. It could be 1 inch to 20 000 inches, 1 cm to 20 000 cm or possibly 1 'little green men' Martian mile to 20 000 'little green men' Martian miles. The reason the ratio is unitless is that a : b can also be written as (a cm/b cm) and then 'cm' cancels. Nevertheless, restoring the units and converting within the metric system makes the problem easier for Mary to visualise.

1 cm : 20 000 cm

1 cm : 200 m, as there are 100 cm in a metre

1 cm : 0.2 km, as there are 1000 m in a km. Of course, Mary could have converted directly from cm to km, using 10 ^ (-2 - 3) = 10^(-5) as her conversion factor.

Now Mary imagines a forest with a rather irregular perimeter.

She remembers placing a leaf on a square grid in Year 6 to estimate its area.

Consequently, she thinks to take a pair of scissors and cut the forest up in little green squares. Each square will be exactly 1 cm long and exactly 1 cm wide. She imagines cutting out 50 such squares, some of which might have to be glued together from various odds and ends.

Each square cm represents 0.04 km2

Thus 1cm2 to 0.04 km2 is the areal ratio of the map.

Algebraically, Mary inputs a ratio into a function machine and gets out a fraction: 1 cm : 0. 2 km → (1cm/0.2 km)
Her next function machine is a squaring machine: (1cm/0.2km) → (1cm/0.2km)2
Her next function machine expands the brackets (1cm/0.2km)2→ 1cm2/ 0.04 km2
Her final function machine converts the fraction back to a ratio: 1cm 2/ 0.04 km2 → 1 cm2 : 0.04 km2.
Please note the arrows. The inputs are not equal to the outputs, so Mary knows it would be wrong to jot down: 1cm/0.2 km = (1cm/0.2km)2.

Since 1 cm2 on the map represents 0.04 km2 on the ground, 50 cm2 will represent 50 times 0.04 = 2 km2 on the ground.
The new map scale uses 8 cm2 to represent the same 2 km2 on the ground. Consequently it uses 1 cm2 to represent 2/8 km2 = 0.25 km2. The areal scale of the new map is 1 cm2 to 0.25 km2. Mary knows she needs to take square roots of both sides to get the linear scale. The square root of 1 cm2 is just 1 cm. The square root of 0.25 km2 is 0.5 km. (While the square root of a counting number is smaller that the original number, it is not the case for fractions. You should recall that the square root of a fraction is larger that the number with which you begin.) Mary now knows the linear ratio of the new map is 1 cm to 0.5 km, i.e.,

1 cm : 500 m
1 cm : 50 000 cm
1 : 50 000

While Mary has been thinking of leaves, pretending to cut up the map and using function machines to convert between linear and areal scales and vice versa, Johnny has been mostly staring out the window and muttering incantations. Suddenly he begins to scribble furiously so that he and Mary annouce -virtually simultaneously- to the teacher, “The ratio for the new map is 1 to 50 000.”

As the class wants to know what Johnny was thinking, the teacher calls him to the white board where he writes:

The new areal map scale, A' to the original areal map scale, A is 8 : 50 or equivalently 16 : 100.

But A = L2 by definition of area and similarly A' = (L')2. I will make these substitutions.

(L')2 : L2 is 16 : 100

However I want L' and L, so I will square root everything.

Sqrt L'2 to sq rt L2 is as sq rt 16 to sq rt 100

Thus L' : L is 4 : 10

Converting the above line to fractional form, L'/L = 4/10 or equivalently L' = 0.4 L.

Given L is 1 : 20 000, consequently L' is 0.4 : 20 000 which is the same as 1 : 50 000. QED.

Conclusion:

While Mary does use algebraic ideas (particulary, function machines) it seems fair to say that her thinking is rather more 'concrete' (or dare I say, grounded) than Johnny's. Vice versa, Johnny's solution is much more abstract than Mary's. But which is better? Certainly, Mary's solution is more accessable to the average student. By contrast, most students appear to believe Johnny has pulled a rabbit from his magician's hat.

## Sunday, September 11, 2011

### Ideal Mixtures

The teacher gives Johnny and Mary’s class a routine problem involving rational numbers.

The Problem: A goblet contains 100 mL of water (H2O) and a bottle contains 100 mL of red wine.  I pour 10 mL from the bottle into the goblet and then 10 mL from the goblet back into the bottle. How much wine is now in the goblet and how much water is now in the wine?

Introduction: If you like wine, you may be wondering who would do that sort of thing.  On the other hand, if you dislike wine, you could be thinking that textbooks have no business referring to alcoholic beverages.   However, these and any number of other non-mathematical concerns will not solve the problem.  Sadly, if you are the sort of person who is easily sidetracked by this sort of speculation you are unlikely to appreciate the usually contrived world of school mathematics.

Instead, the following rather technical suppositions are more useful:
·         Volumes of water and wine are strictly additive. The total volume always remains 200 mL.
·         Red wine cannot change into water and vice versa.
·         On contact, red wine and water mix completely and cannot be unmixed. Thus, they must always be transferred proportionally.
·         milli-Litres, mL, of water or wine may be subdivided into micro-Litres, μL, and they in turn may be subdivided into nano-Litres, nL, and so on indefinitely…

A Note about Notation: The notation (W, R) will be used throughout.  For example (90, 9) means 90 mL of water and 9 mL of red wine.

A First Attempt: The teacher asks, “What is wrong with the proposed solution, below?”
 Goblet 10 mL spoon Bottle Initial condition (100, 0) (0, 0) (0, 100) During the 1st transfer (100, 0) (0, 10) (0, 90) After 1st transfer (100, 10) (0, 0) (0, 90) During the 2nd transfer (90, 10) (10, 0) (0, 90) x

Mary explains that the second transfer is “illegal” as the ratio 90 : 10 is obviously not the same as 10 : 0.  The first ratio simplifies to 9 : 1.  The second ‘simplifies’ to 1 : 0 which some brave souls might perhaps re-write as : 1.  (As a practical matter, from the 100 : 10 mixture, how is it that we only select 10 parts of water to be transferred?)

A Second Attempt: Johnny suggests the following table, instead.

 Goblet 10 mL spoon Bottle Initial condition (100, 0) (0, 0) (0, 100) During the 1st transfer (100, 0) (0, 10) (0, 90) After 1st transfer (100, 10) (0, 0) (0, 90) During the 2nd transfer (90, 9) (10, 1) (0, 90) x After 2nd transfer (90, 9) (0, 0) (10, 91)

The ratio for the second transfer is correct since 90 : 10 <=> 10 : 1.  However, the teacher reminds Johnny that he has fallen afoul of one of the “givens” of the word problem, pointing at: “and then 10 mL from the goblet back into the bottle.”  With the hint, Johnny 'remembers' that the 2nd transfer was a total of 11 ml.

The Proposed Solution: Mary draws up the following table.

She exclaims, “Look, there is just as much water in the wine as there is wine in the water!”

Is the solution correct?: While the solution is very elegant, Johnny wants to argue that it is not absolutely correct. In Chemistry class, he has been told there is a smallest piece of water, termed a molecule.  Therefore, he argues we cannot write ‘90.909090 dot dot dot.’  Instead, at some point the decimal must truncate. It could be after 20 decimal points or 53 or some other number, but thereafter the pattern would be broken.
“Equivalently”, Johnny says, “We can divide 99 by 11 but we cannot divide 100 by 11.  Mary can ‘smooth over’ that difficulty by converting 100 mL to 100 000 μL. Nevertheless, as that too will still not divide by 11, she must make a further change to 100 000 000 nL.  But neither will that divide by 11.  In typically human fashion she simply made the problem recede, not vanish.   Eventually we must reach a smallest volume that contains a single molecule of water, forcing us to stop the process of ‘converting between units’.”

Are any of the solutions reasonable?: Mary appeals to the teacher, who confirms that for the purpose of IGCSE Mathematics, her answer is correct, as it is mathematically more sophisticated.  But he reminds her that it is acceptable to round answers to 3 d.p., thus, (90.9, 9.09) and (9.09, 90.9).

The teacher further explains that Mathematics is an idealization of reality. In every problem, there is always a translation (or idealization) process from reality to mathematics, in which simplifying assumptions are made.  If we do not simplify enough the problem may be insoluble.  However, if we oversimplify, the conclusion we draw may be invalid.  For IGCSE Mathematics these concerns are generally pushed to the background as they belong more to a Philosophy or to a “Theory of Knowledge” course and our focus is generally on ensuring students possess accurate and sophisticated methods of calculation.

Footnotes:
1) In the kitchen (but not on the exam page), if I transfer the water and the wine by a teaspoon, I am exceedingly unlikely to be able to distinguish between 10 mL and 11 mL. Thus, for many practical purposes Johnny’s attempted solution is good enough.
2) Johnny has continued to calculate and tells the class that 1 mL of water is 30 000 000 000 000 000 000 000 molecules.  (Or near enough.)  Consequently, a nanolitre is still an amazing 30 000 000 000 000 molecules.  He now agrees that in leaving aside the dot dot dot, you can still conclude that the ratio of water to wine in the goblet is the same as the ratio of wine to water in the bottle.
3) A chemist who reads this blog may want to dispute the additivity of volumes of water and wine. (S)he will say 100 ml of water mixed with 100 ml of red wine actually makes less than 200 ml of solution! (But that is a story for another day).