Good! Let me reiterate the point. The idea behind the Remainder Theorem is that if you just want to know the remainder R when a polynomial f(x) is divided by (x − a), then you don't need to do any long division; you can directly substitute x = a so that f(a) = R.

Factor Theorem states if R happens to be zero: R = 0, then (x − a) is a linear factor of f(x).

In practice, you really don't need to find the first linear factor. Because CASIO fx-570/911 has a built-in function EQN to find the roots of a polynomial (up to 3rd degree), you'll find either 3 unequal real roots, or 3 real roots & at least 2 are equal, or 1 real root & 2 complex conjugates.

In fact, the factorization of cubic polynomials is a math skill, pretty much the same as you learned the factorization of quadratic polynomials in Form 2. In exam, you'll probably encounter a question like the following:

Given a cubic function, f(x) = x³ − 6x² − 7x + 60

(a) Find the remainders if f(x) is divided by i. (x − 1), ii. (x − 2), iii. (x − 3), and iv. (x − 4). ... [2 marks]
Long division is NOT required. Just find f(1), f(2), f(3), f(4).

(b) Using the results from (a), solve the equation x³ − 6x² − 7x + 60 = 0 ... [4 marks]
Do a long division for 2 marks and solve the quadratic part for another 2 marks.

The image size and resolution should be good enough for reading. Are you using a Desktop? Anyhow, please download the pdf version.  Tutorial_1_Remainder_Theorem.pdf ( 194.27k ) Number of downloads: 18


This post has been edited by Critical_Fallacy: Nov 23 2013, 08:02 PM

Could you show me the inspection method, please?

Wow! Pretty fast calculations. Thanks for showing me the Mental Inspection method. Although Long Division is the official method, in practice, we use Synthetic Division because it is more systematic and has fewer steps. Perhaps, you wanna have a look in Part 2 of the Partial Fraction Tutorial.

This post has been edited by Critical_Fallacy: Nov 25 2013, 11:39 AM

Hi maximR, this is the tutorial for Partial Fractions, which split into 4 parts. You will become a Master of Partial Fractions in 60 minutes.

Part 1: Introduction to Partial Fractions

Part 2: Rules of Partial Fractions & Synthetic Division

Part 3: Denominators with Irreducible Quadratic and Repeated Factors

Part 4: Other interesting methods :: Differentiation, Residue, and “Cover-up”

Pre-print version (pdf):  Tutorial_2_Partial_Fractions.pdf ( 205.06k ) Number of downloads: 17

Vacuum Cleaner
-------------------
As the fan blades of the Vacuum Cleaner turn, air particles are driven forward. This causes the air pressure decreases behind the fan. This pressure drop behind the fan creates suction, a partial vacuum, inside the vacuum cleaner. The ambient air pushes itself into the vacuum cleaner through the intake port because the air pressure inside the vacuum cleaner is lower than the pressure outside. Although we can associate Bernoulli effect at the intake port, the suction mechanism is just like the pressure drop in the straw when you sip from your drink.

Bunsen Burner
-------------------
When the Bunsen Burner is connected to a gas supply, the gas flows at high velocity through a narrow passage in the burner, creating a region of low pressure. According to Bernoulli's Principle, the outside air, which is at atmospheric pressure, is drawn in and mixes with the gas. The mixture of gas and air enables the gas to burn completely to produce a clean, hot, and smokeless flame.

If you ask me, the Bernoulli's Principle has a more significant effect on the operation of the Bunsen Burner.

Before jumping to the Explanation part, let's review the basics. The following is a short lecture on Electric flux.

By definition, Electric flux (Φ) is proportional to the number of electric field lines penetrating some surface. The total number of lines penetrating the surface is proportional to the product of the magnitude of the electric field (E) and surface area (A) perpendicular to the field:

Φ = EA

If the same surface under consideration is not perpendicular to the field, where the normal to the surface of area A is at an angle θ to the uniform electric field E, the flux through it must be less than that given by the above equation, and a modified equation is derived:

Φ = EA cos θ

From this result, we see that the flux through a surface of fixed area A has a maximum value EA when the surface is perpendicular to the field (when the normal to the surface is parallel to the field, that is, when θ = 0° in Fig. 24.2); the flux is zero when the surface is parallel to the field (when the normal to the surface is perpendicular to the field, that is, when θ = 90°).



In more general situations, the electric field may vary over a large surface. Therefore, the definition of flux given by the Modified Equation has meaning only for a small element of area over which the field is approximately constant. Consider a general surface divided into a large number of small elements, each of area ΔA.



It is convenient to define a vector ΔAi whose magnitude represents the area of the ith element of the large surface and whose direction is defined to be perpendicular to the surface element as shown in Figure 24.3. The electric field Ei at the location of this element makes an angle θi with the vector ΔAi. The electric flux ΔΦi through this element is

ΔΦ = Ei ΔAi cos θi = Ei•ΔAi

where we have used the definition of the scalar “Dot” product of two vectors. Summing the contributions of all elements gives an approximation to the total flux through the surface:

Φ ≈ Σ(Ei•ΔAi)

Using Calculus, if the area of each element approaches zero (ΔA → 0), and the number of elements approaches infinity (i → ∞), then the sum (Σ) can be replaced by an integral (∫). Therefore, the general definition of electric flux is

Φ = ∫ Ei•dAi

The above general equation is a surface integral, which means it must be evaluated over the surface in question. In general, the value of Φ depends both on the field pattern and on the surface. We are often interested in evaluating the flux through a closed surface, defined as a surface that divides space into an inside and an outside region so that one cannot move from one region to the other without crossing the surface.



For a closed surface in an electric field, the area vectors are normal to the surface and by convention, always point outward. Therefore the electric flux Φ where the electrical field lines:
(1) crossing a surface area element from the inside to the outside and 0° ≤ θ < 90° is positive,
(2) grazing a surface area element (perpendicular to the vector ΔAi or parallel to the surface) and θ = 90° is zero,
(3) crossing a surface area element from the outside to the inside and 90° < θ ≤ 180° is negative.

The net flux through the surface is proportional to the net number of lines leaving the surface, where the net number means the number of lines leaving the surface minus the number of lines entering the surface. If more lines are leaving than entering, the net flux is positive. If more lines are entering than leaving, the net flux is negative.

Explanation: Total flux that passes through the cylindrical Gaussian surface:


This post has been edited by Critical_Fallacy: Nov 27 2013, 01:56 PM

Maybe you missed the 2nd and 3rd paragraph. Referring to Fig. 24.2, when the surface A is parallel to the E field, what do you think the angle θ would be?

θ = 90°

Isn't it? Now, by applying the modified formula, we have

Φ = EA cos 90°

Since cos 90° = 0 (in trigonometry), whatever number multiplied by 0 is 0, then

Φ = EA × 0 = 0

Therefore, the flux is zero when the surface is parallel to the field. Got it now?

Putting it another way, imagine when the surface is parallel to the field, the E lines do NOT hit the surface area A, do you see that? Since no hitting, then the flux is zero, by definition stated on the 1st paragraph.

This post has been edited by Critical_Fallacy: Nov 27 2013, 05:01 PM

Let's apply Polya’s four-step problem-solving process in this case.

STEP 1: Understand the problem

Suppose that the cubic equation f(x) = ax³ + bx² + cx+ d = 0 has roots α, β, γ.

f(x) = ax³ + bx² + cx+ d = 0

x = α, β, γ

Find a new cubic equation g(x) with the roots h(α); h(β); h(γ).

Given: x = h(α); h(β); h(γ)

Find: g(x) = 0

STEP 2: Develop a plan to solve the problem

Remainder and Factor Theorems tell us that: f(α) = f(β) = f(γ) = 0

Since g(x) = 0, we can apply the law of conservation of energy and we pick: g(x) = f(γ) = 0

Both sides must contain x. And the link between x and γ is x = h(γ). Manipulate it: γ = h^-1(x)

STEP 3: Carry out plan

g(x) = f(γ) = 0

g(x) = f[h^-1(x)] = 0

g(x) = px³ + qx² + rx+ s = 0

STEP 4: Look back and check the answer.

The roots of the cubic equation px³ + qx² + rx+ s = 0 should be exactly the same as h(α); h(β); h(γ).

By the way, this sorcery is called “by means of a substitution”, and the substitution is γ = h^-1(x) = 2x / (x − 1). Although this problem is inherently A-level / STPM level, it can be solved using SPM Form 4 Add Math Functions knowledge + Standard Algebraic Manipulations (SAM). Thanks ystiang for he has done a good job!

This post has been edited by Critical_Fallacy: Nov 28 2013, 11:10 PM

Thanks Dr v1n0d! Please be a “Mathematician by day” in this thread so that you can provide some guidance to students if they need help, when I'm not around.

You're welcome, and you are gifted, YiJun. Please help some students to understand the physics principles and concepts when I'm not around.

My Critical Story: When I was younger, I asked my SPM Modern Math teacher “What is the value for log(−1)?” but he told me that the logarithm of a negative number does not exist! Well, I was skeptical and at 16yo, it was possible for me to think log(−x) = log(x), so log(−1) = log(1) = 0. My mathematical proof was the following:

Since (−x)*(−x) = x², therefore log[(−x)*(−x)] = log(x²) = 2*log(x).
But log[(−x)*(−x)] = log(−x) + log(−x) = 2*log(−x), so I got log(−x) = log(x).

I still remember the look on his face, and he dismissed my thought without further explanation, although I was algebraically correct.

Hi ailing tan, dividebyzero & maximR, this is my blood, sweat and tears for Complex Numbers, which is a breathtaking 14-part tutorial notes. Once you learn the fundamental examples, you will be able to do almost every Complex Numbers problems in A-level/STPM tests, because such nominal tests can't be recycled as much, or derivable from over 30 fundamental examples.

You are probably aware that Complex Numbers have something to do with √(−1). In fact, Complex Numbers enable engineers, mathematicians, & scientists to solve many Insoluble Problems beyond the Real Numbers domain. When you have completed this tutorial, you will be able to find the value of log(−1). And you will become a very “Complex Person” in 7 hours.

Prerequisites: SPM Arithmetic, Algebra, Geometry, Indices, Logarithms, Trigonometry, Binomial expansions

Part 1: Introduction to the symbol i, Powers of i, & Complex numbers (z)

Part 2: Addition & Subtraction, Multiplication, Complex conjugates, Division

Part 3: Equal Complex numbers, Graphical representation, Polar form

Part 4: Exponential form of a Complex number

Part 5: Exponential form of a Complex number (cont'd)

Part 6: Tricks, Powers of Complex Numbers, Roots of Complex Numbers

Part 7: Exploiting the 6th Rule of Law of Indices

continue to Post #122...

This post has been edited by Critical_Fallacy: Nov 30 2013, 09:00 PM

continue from Post #121...

Part 8: Multiple examples of Powers & Roots of Complex Numbers

Part 9: Trigonometry and Complex Numbers

Part 10: Trigonometry and Complex Numbers (cont'd)

Part 11: Multiple examples of Trigonometry problems

Part 12: Loci Problems

Part 13: Loci Problems (cont'd)

Part 14: Exponential and Trigonometric Functions of z, Logarithm and General Power of z

Pre-print version:  Tutorial_3_Complex_Numbers.pdf ( 537.3k ) Number of downloads: 32


This post has been edited by Critical_Fallacy: Nov 30 2013, 04:31 AM

Your approach is basically correct. The mistake probably occurred at the Numerator of the Sum: 1 − z^(N − 1). When you add up the terms up to the last term n in a converging geometric series, the Sum formula should compute the sum of n+1 terms (compare the LHS and the RHS of the equation below).



Another mistake occurred at the denominator part when you multiplied two complex conjugates to produce a real number. Mathematicians expand to proper polar forms when doing the complex multiplication. The best practice is to avoid writing shorthands of polar form r∠θ when doing arithmetic operations, even though the work is tedious.

To find out how to solve the rest of the problem, please check out the Examples 23 & 24 (found in Parts 11 & 12 of my Complex Number tutorial note). I believe they would be helpful, because your problem is exactly derivable from the fundamental examples.

As a smart student, choose the selective question fastidiously, because complex trigonometric proof often requires one to go the extra mile and then return. For example, in proving the cos part (Real), the student inevitably has to expand the sine part (imaginary) as well, even though it is not required to be proven. For solving a 10-mark question, you need the 20-mark worth time. One thing good is that most A-level/STPM complex problems cannot run away far from the fundamental examples.

This post has been edited by Critical_Fallacy: Nov 30 2013, 04:10 AM

"Knock-Knock" anybody's home? Luck is at doorstep...



Tip: Complex conjugates are a pair of complex numbers, both having the same real part, but with imaginary parts of equal magnitude and opposite signs. For example, The conjugate of (−3 + 4i) is identical to it, except for the opposite sign of the imaginary part: (−3 − 4i).

This post has been edited by Critical_Fallacy: Dec 1 2013, 01:23 AM

SPM question? Simple but you need three evanescent qualities as shown below:

(1) Intelligence.
You see √ on the LHS. But you cannot get over to RHS to assess a and b directly. You are walking into a trap. So, how do you put both sides on equal ground using algebra? +, −, ×, ÷, ^, √? Which is the one?

(2) Originality.
If you think you have found a and b, you just fell into the trap! When students deal with quadratic stuffs, they tend to ignore the original unwritten sign “+” in front of the “beg-to-be-found” √. Who’s fault? Not me. It’s the mathematicians’.

(3) Excellence.
The more elegant your answer is, the more excellent you are. In fact, There are two aspects to mathematics:“discovery and proof” ... and they are of equal importance. We must discover something before we can attempt to prove it, and we cannot be certain of its truth until it has been proved.

This post has been edited by Critical_Fallacy: Dec 2 2013, 08:48 AM

In fact, the power of 2 is rather ambiguous, because it has a double meaning.

(−3)² = 9

(+3)² = 9

The question from Shanks is designed to trick those “I-think-I'm-smart” people. Most instantly find a = 81 and b = 88.

If we examine the equation carefully, the outer √ on the LHS suggests that (√a − √b) > 0 on the RHS.

This post has been edited by Critical_Fallacy: Dec 2 2013, 03:03 PM

Sorry, I cannot explain that. But I can tell you are intuitively correct.

y = 2*cos(3x-4)

Using chain rule, you'll get

dy/dx = 6*sin(4-3x)

∂soo,

That's an obvious typo. Again, you are intuitively correct.

Are you learning Al-level / STPM Differentiation right now?

The orange box is critical. The blue boxes are non-critical comment but they are worth pointing out. You could prove by mathematical induction. Remember the last quality: Excellence?

To a strict math marker, you are plucking the figures (happen to be the same as answers) out of thin air, because those workings are illegitimate.

I'm wondering, from which chapter of the textbook did you find that Hot Air Balloon Physics question? That's because the principles behind hot air balloon physics are usually related to the chapter. At first glance, it probably has something to do with the Archimedes’s Principle (Buoyant Force).

Since, the temperature of the air (100 °C), volume of the envelope (3000 m³), and the unwritten atmospheric pressure (1 atm) are known, it seems to relate to the equation of state for an ideal gas: PV = nRT.

Perhaps, based on the above principles, you can work out something and post up here for further discussion.