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Maths Core 2 ...something nice and easy to start the topic... Whenever you are given a fraction with a single term appearing in the denominator, which appears in the numerator, that term can be cancelled (obviously). For example: = 3x³  2x² + 5. Easy. It gets a little harder with a term like (x2) on the bottom though, but that will usually require you factorising the top. E.g.: = = (x2) Hint: You know you've got it right because the denometer in this kind of equation is usually one of numerator terms. Slightly trickier is when you have to factorise both top and bottom first, but it's still easy enough. :) For example: = = All pretty easy, and should remind you of GCSE. Long Division Now it gets interesting. Ever wondered how to factorise a x³ equation? With great difficulty... But here's the first of a few methods that will allow you to factorise any polynomial equation. (polynomial means 'many terms'. x + x³ + x² + x + 1 for example.) Remember WAAY back in primary school how they taught that horrible long division method? Well, it's back with a vengeance. A reminder with numbers to start with: The answer you get above is called the quotient. If you have a remainder, this is just called the remainder. Now, that method may be extremely longwinded, but it is the method you need to know for solving algebraic divisions. Take the following example: The first step is to divide x^{3} by x, to get x^{2} above the line. Then below, you times the x^{2} by x  4.
Next, take the x^{3}  4x^{2} from x^{3}  5x^{2}. If you're doing it right, the first two should cancel. Remember your brackets and  sign. Because we have 4x^{2}, we end up adding 4x^{2}. Write the answer below: Then bring down 5x. Divide the x^{2} by x to get x and write this above. Times x by x4 and write the answer below. Take (x^{2} + 4x) from (x^{2} + 5x). Be careful with the minuses! Here, if you're doing it right, the x^{2} should cancel. Write the answer below and then bring down 4. Divide the x by x, and write 1 above. Finally times 1 by x4 and write below. This will cancel and leave 0. You have reached the answer. When x^{3}  5x^{2}  5x  4 is divided by x  4 the quotient is x^{2}  x + 1. Unfortunately, not all questions they might give you in the exam will be nice and divide exactly. In that case, you'll get a remainder, and you'll need to be even more careful with your working out. The questions might also miss out an x term. For example, in: (x^{3}  2) (x  3), there's no x^{2} or x term. You need to write these into the division with 0x^{n} and thus get: From there, you just use the same method as before to solve: and from that you get a quotient of x^{2} + 3x + 9 and a remainder of 25. From this we know that x^{3}  2 is the same as: (x^{2} + 3x + 9)(x  3) + 25. Factor Theorem The nice thing about maths is that when you know the hard method, you can find the easier one. For algebraic division, we use something called factor theorem and remainder theorem. Factor theorem: For any polynomial f(x), if f(a)
= 0, then (x  a) is a factor of f(x). This gives us the ability to find the factors of polynomials or prove them. Show that (x  2) is a factor of the polynomial (x^{3}  2x^{2} + x  2). Using factor theorem we know that when f(a) = 0, (x  a) is a factor. Therefore, we can say that f(2) = 0, if it's a factor. Now put 2 in for the values of x: f(2) = 2^{3}  2x2^{2} + 2  2 Therefore: f(2) = 8  8 = 0 Because it equals 0, we know (x  2) is a factor. You can use trial and improvement to find the factors: Factorise 2x^{3} + 3x^{2}  32x + 15 completely. Let: f(x) = 2x^{3} + 3x^{2}  32x + 15
To factorise completely, you then divide (2x^{3} + 3x^{2}  32x + 15) by (x  3). It should come out with no remainder  a little proof to check your factor is right: From this you can say: (x  3)(2x^{2} + 9x  5) = 2x^{3} + 3x^{2}  32x + 15 Factorise the quadratic to get an answer of: (x  3)(2x  1)(x + 5). Remainder Theorem Finally, the last part of simplifying equations is remainder theorem. After learning to use factor theorem, this is easy. If f(x) is divided by (x
 a), then f(a) gives the remainder. What's the remainder when f(x) = x^{3} +4 is divided by (x  2)? So f(x) = x^{3} + 4 => f(2) = 2^{3} + 4 2^{3} + 4 = 12, so the remainder is 12. Using this principle, the exam might ask a slightly harder version: f(x) = ax^{3} + bx^{2}  13x + 6. When f(x) is divided by (x + 1) the remainder is 18 and (2x  1) is a factor of f(x). Find the values of a and b. First, it's a case of writing in what you know: For (x + 1) we get:
For a factor of (2x  1) we get:
From here, it's a case of simultanious equations to solve:
Sine and Cosine Rules More stuff that should be vaguely familiar from GCSE... To fine a side using the Sine Rule, you use the following formula: For an angle, it's simply the inverse: (Capital letters for the angles, lower case for the sides ALWAYS.) To illustrate, take the following shape:
To find y, you need two parts of the sine rule: Put numbers in and manipulate and solve the equation to get:
Special cases of Sine might have two solutions. (The exam question will usually hint towards this if it says find the solutions.) Here's what it might looks like: For this to happen, the given angle (30° in the example) must be smaller than the two angles we're finding.
Cosine Rule The formula for finding a side is: a² = b² + c²  2bc.CosA. For an angle, it's: If the
angle is acute, CosA is negative. Find b in the following triangle:
Find the smallest angle in the following triangle: Note: The smallest angle is always opposite the smallest side. An easier way that finding every angle! Remember, to remove Cos from C, just do the inverse of Cos. Area of a circle This simple formula lets you work out the area of any circle. Pretty cool:
Exponentials and Logarithms What is an exponential function? Well, you might recall the graph from GCSE...
For example, the graphs of y = 2^{x} and y = 2^{x}: y=2^{x} y=2^{x} As you can see, the minus is a reflection in the yaxis. (not surprising, if you consider Core 1 Tranformations). While the lines get very close, they never cross the x or yaxis. Therefore, the x and yaxis are asymptotes. Logarithms Another strange maths word... Look at the following: 2^{5} = 32 This is written out in standard form. Another way of writing it would be in log form: 5 = log_{2}32 Both mean exactly the same. However, because your magic calculator can work out log_{10} numbers, logarithms become very useful. Before we get ahead of ourselves, let's explain what each bit means...
In general:
Three important rules for manipulating logs:
IMPORTANT NOTE: These only work when the bases are the same. Proofs:
Examples: Simplify: log_{a}8  log_{a}4 + log_{a}5
So, it's all very well and good messing around with logs, but how can we USE them? It's quite simple really with the calculator. Take the following example:
How can you figure out what x is? Well, you could use trial and error, or you could convert it all to base 10 form (which we just write as log):
Next, use the laws we know already:
Finally, stick that into your calculator and out pops the answer:
Solve: 6^{(2x+1)} = 3^{x}
Trigonometric Functions The Special Trig Angles: You MUST learn the following table:
