Mathematics GCSE (1)

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prime factorism

"Prime Factorization" is finding which prime numbers multiply together to make the original number.

Example 1: What are the prime factors of 12 ?

It is best to start working from the smallest prime number, which is 2, so let's check:12 ÷ 2 = 6

Yes, it divided evenly by 2. We have taken the first step!

But 6 is not a prime number, so we need to go further. Let's try 2 again: 6 ÷ 2 = 3

Yes, that worked also. And 3 is a prime number, so we have the answer:12 = 2 × 2 × 3

 As you can see, every factor is a prime number, so the answer must be right.

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multiplication of surds

Manipulations of surds

There are two basic identities you need to know.

  • \sqrt{a} \times \sqrt{b} = \sqrt{ab} (http://www.thestudentroom.co.uk/latexrender/pictures/30/303e11f32355b3afe5b6e690c7651776.png). For example, \sqrt{2} \times \sqrt{3} = \sqrt{6} (http://www.thestudentroom.co.uk/latexrender/pictures/d9/d9c6c436e4145f3902bd18c86f510aba.png).
  • \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} (http://www.thestudentroom.co.uk/latexrender/pictures/9e/9ef27578a2b8028bf4ca2d4ab61e9895.png). For example, \frac{\sqrt{20}}{\sqrt{5}} = \sqrt{\frac{20}{5}} = \sqrt{4} (http://www.thestudentroom.co.uk/latexrender/pictures/c5/c5a450a9f86d9e417f76a478566d1901.png)

IMPORTANT: There are no simple identities for adding and subtracting surds - in most cases, something like \sqrt{2}+\sqrt{5} (http://www.thestudentroom.co.uk/latexrender/pictures/26/26cc507ceec187c8acf95bd924ec97bb.png) can't be simplified!

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rationalising the denominator

Rationalising the denominator

This is the tricky bit. When you're dealing with fractions, Edexcel hates it when you leave surds on the bottom - you have to "rationalise the denominator". For example:

\frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5} (http://www.thestudentroom.co.uk/latexrender/pictures/19/1994768eae799a51cf5d7835dd859e95.png)

See what we did there? We wanted to get rid of the surd from the denominator, so we multiplied top and bottom by the surd. That's basically it. Another example:

\frac{1}{2\sqrt{5}} = \frac{1}{2\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{2 \times \sqrt{5}^2} = \frac{\sqrt{5}}{10} (http://www.thestudentroom.co.uk/latexrender/pictures/53/53dd0ca651f793a834a32def04131b7e.png)

There are some fractions where this method won't work, though, because the denominator has more than one term in it. In that case, we have to use the difference of two squares. Take a look:

\frac{1}{\sqrt{2}+\sqrt{5}} = \frac{1}{\sqrt{2}+\sqrt{5}} \times \frac{\sqrt{2}-\sqrt{5}}{\sqrt{2}-\sqrt{5}} = \frac{\sqrt{2}-\sqrt{5}}{\sqrt{2}^2 - \sqrt{2}\sqrt{5} + \sqrt{2}\sqrt{5} - \sqrt{5}^2} = -\frac{\sqrt{2}-\sqrt{5}}{3} (http://www.thestudentroom.co.uk/latexrender/pictures/9b/9b2a18dc47ac94a9be906a148ed4c1c2.png)

The irrational bit cancels, making life easier for us all!

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Comments

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Conor

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Conor

Matthew you curly freak :P

matthew heathcote

ye mate

Mr d

bit short

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