Hi maximR, this is the Solution of Cubic equations & Quartic equations. Most likely you will master this in about 10~15 minutes.



You have already learned the basic factorization skill in Form 2 and quadratic solution by completing the square & formula in Form 4 Add Math. So, tell me now, are your Math skills sufficient to solve cubic equations and quartic equations with the systematic procedure?

This post has been edited by Critical_Fallacy: Nov 23 2013, 02:20 AM

Sorry I went offline yesterday , it was quite late and I had to go to bed . Anyway , I'm done with Remainder Theorem and Factor Theorem .

As for your last question , is my reasoning correct ?

Since the remainder theorem states that if f(x) is divided by (x-a) , then the remainder = f(a) and factor theorem states that if f(x) is divided by (x-a) and it equals zero , then (x-a) is a linear factor , thus , if x = a , then :

a^3 - 6a - 7a + 60 = 0

Factoring using my calculator , since I haven't learned Cubic and Quartic Eqn ( can you please re-post the tutorial ? it's too small for my eyes ) yields :

(a+3)(a-5)(a-4) = 0

Therefore :

(x+3)(x-5)(x-4) = 0

Thus , I've found the linear factors of the function without using the first linear factor . I hope it's correct .

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

An easier method would be by inspection, which should be faster than long division. But long divison is safer.

This post has been edited by VengenZ: Nov 23 2013, 08:00 PM

Could you show me the inspection method, please?

Its somewhat same with long division,
for example we know that (x-4) is one of the factor for the eqn x³ − 6x² − 7x + 60

so x times x² will get you x³. we will get
(x-4)(x² + bx +c)

we know that -15 times -4 is equal to 60
we get
(x-4)(x² +bx -15)

heres the kinda tricky part,
-15 * x = -15x
to get -7x, we know that -15x must be added with 8x, and we also know that -4 * -2x = 8
so we get (x-4)(x² -2x -15), when factorised completely, we'll get (x-4)(x+3)(x-5).

This method saves time as we can just look at the equation and factorise it straight away with some mental calculations.

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

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.

Actually , I wanted to tell you that the day after you posted that Cubic/Quartic tutorial , I'd read it and it's safe to say I'm done with it . Maybe more practices so I can be faster with algebraic long division .

I'll look through these tutorials probably after my Biology paper . Thank you so much , someone who spends his time creating tutorials for a complete stranger deserves an award . Don't really know how to re-pay you .

hi i got a question here. kindly refer to the photo attached. why the total flux flow thru both ends are zero??? is it beacuse of sin0??? why not cosine 0 ??? when to use cos0? and sin0???


Attached thumbnail(s)

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

Excellent thread, keep 'em coming!

hi i don't understand why the flux is zero when the surface is parallel to the field ?

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

yeah i got it now! thanks for your explanation again! your explaination is better than the book !

Hi everyone,
I would first introduce myself. My name is YiJun, 19 and I'm studying STPM in MBSSKL.
I participated the IPhO training in UKM a few months ago, not sure whether did I met any of you.
After the exams and university applications I would started reading some Olympiad math and physics materials, perhaps we can share knowledge with each others. Nice to meet you

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

Thumbs up for both of you. I kept trying x = y/(y-2) and even (x-a)(x-b)(x-c)=0 to solve and failed superbly.

Now, another math question appears.


So far I have reached here. How can I execute these highlighted areas?


This post has been edited by dividebyzero: Nov 29 2013, 10:06 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.