## 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.