Title: 2. Factorisation of whole numbers
Course: Demonstrating STACK (v4.3.7)

Factorise \([value]\) into primes:
Write your answer as powers of primes, e.g. \(p_1^{n_1}\times\cdots\).
Feedback: To factorise a number, \(n\) into primes the best trick is to divide by the smallest prime, \(2\). If the resulting value, \(m\), is an integer then we write it as \(2 \cdot m\). If not, then we try the next largest prime, \(3\), and repeat the process until we have obtained a product of the form \[p_1 \cdot p_2 \cdot ... \cdot 1, \] where p are prime numbers. In this particular example: \[ [value] = [value]. \]

Reduce the following fraction to it's lowest terms: \[\frac[value][value].\]
Feedback: Factorise the numerator and the denominator into prime factors, then cancel matching factors. In this example: \[ \frac[value][value] = [value]. \]

Find the Highest Common Factor (hcf) of \([value]\) and \([value]\).
Feedback: There are multiple ways to find the HCF. One way is to use Euclidean Algorithm. However, a quicker method, if the numbers are small is to factor the two numbers into primes. In this example: \[ [value] = [value], \] \[ [value] = [value]. \] Then you take the common factors and multiply them together to obtain [value] which must be the highest factor common to both [value] and [value].

Find the Lowest Common Multiple (LCM) of \([value]\) and \([value]\).
Feedback: To find the LCM of two numbers first factor into primes: \[ [value] = [value], \] \[ [value] = [value]. \] Then we define the LCM as the product of the ALL the primes in both \([value]\) and \([value]\) (However if a prime appears in both \([value]\) and both [value] then we only count it once.) By doing this we obtain the LCM, \([value]\).

Convert the following mixed fraction into an improper (single) fraction:
\[ [value]\frac[value][value]. \]
Feedback: The form \(a\frac{b}{c}\) is really just short for \(a+\frac{b}{c}\). So we're just adding an integer to a fraction. In order to do this we multiply the integer, \(a\), by the denominator, \(c\), and add it to the the numerator of the fraction. This can be seen from the general rule: \[ a\frac{b}{c} = \frac{ac+b}{c}. \] Therefore in this example: \[[value]\frac[value][value] = \frac{[value]\cdot[value]+[value]}[value] = [value]. \]

Convert [value] into a mixed fraction:

Note, a mixed fraction might be written \(a\frac{b}{c}\) but you have to type: "\(a+b/c\)".
Feedback: \[ [value] = [value]. \]

Rationalise the following surd:
\[\sqrt[value] \cdot \sqrt[value].\]
Leave your answer in the form: \(a \cdot \sqrt{b}\) where b cannot be factored into any square numbers.
Feedback: Firstly use the fact that for positive numbers \(a\) and \(b\): \[ \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}.\] Then factor \(a \cdot b\) into powers of primes. Even powers of primes can be taken outside the square root. In this example: \[ \sqrt[value] \cdot \sqrt[value] = \sqrt{[value]\times [value]} = \sqrt[value] = [value].\]