# How do you factorise these please? 18x^3+9x^2-2x-1;" "2x^3-432;" "6n^4-11n^2-2:" "n^4-1

Nov 21, 2016

The types of factoring are:
Take out a common factor. or take out a common bracket
Grouping
Difference of squares
Sum or Difference of cubes

#### Explanation:

$18 {x}^{3} + 9 {x}^{2} - 2 x - 1 \text{ } \leftarrow$ there are 4 terms, group them in pairs

=$18 {x}^{3} - 2 x + 9 {x}^{2} - 1 \text{ } \leftarrow$ try to place a positive term third

=$\left(18 {x}^{3} - 2 x\right) + \left(9 {x}^{2} - 1\right) \text{ } \leftarrow$ factor each pair.

=$2 x \left(9 {x}^{2} - 1\right) + \left(9 {x}^{2} - 1\right) \text{ } \leftarrow$ common bracket

=$\left(9 {x}^{2} - 1\right) \left(2 x + 1\right) \text{ } \leftarrow$ difference of squares

=$\left(3 x + 1\right) \left(3 x - 1\right) \left(2 x + 1\right)$

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$2 {x}^{3} - 432 \text{ } \leftarrow$ common factor of 2

=$2 \left({x}^{3} - 216\right) \text{ "larr}$ difference of cubes

=$2 x \left(x - 6\right) \left({x}^{2} + 6 x + 36\right)$

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$6 {n}^{4} - 11 {n}^{2} - 2$
Find factors of 6 and 2 which subtract to make 11.
Note that the biggest product of factor is $6 \times 2 = 12$

=$\left(6 {n}^{2} + 1\right) \left({n}^{2} - 2\right)$

=$\left(6 {n}^{2} + 1\right) \left(n + \sqrt{2}\right) \left(n - \sqrt{2}\right)$
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n^4-1)" "larr difference of squares

=$\left({n}^{2} + 1\right) \left({n}^{2} - 1\right) \text{ } \leftarrow$ difference of squares

=$\left({n}^{2} + 1\right) \left(n + 1\right) \left(n - 1\right)$
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HOPE THESE HELP