# How to factorise 2a^6-19a^3+24?

Oct 5, 2017

$\left(2 {a}^{3} - 3\right) \left(a - 2\right) \left({a}^{2} + 2 a + 4\right)$

#### Explanation:

let u=a^3

we have then

$2 {u}^{2} - 19 {a}^{3} + 24$

a quadratic in ${a}^{3}$

$2 \times 24 = 48$

factors of $48$ that sum to $- 19 \rightarrow - 16 , - 3$

we have

$2 {u}^{2} - 3 u - 16 u + 24$

$= \left(2 {u}^{2} - 3 u\right) - \left(16 u - 24\right)$

$= u \left(2 u - 3\right) - 8 \left(2 u - 3\right)$

$= \left(2 u - 3\right) \left(u - 8\right)$

substituting back

$\left(2 {a}^{3} - 3\right) \left({a}^{3} - 8\right)$

the second bracket is difference of cubes

$\left(2 {a}^{3} - 3\right) \left(a - 2\right) \left({a}^{2} + 2 a + 4\right)$

Oct 5, 2017

$\left(a - 2\right) \left({a}^{2} + 2 a + 4\right) \left(2 {a}^{3} - 3\right)$

#### Explanation:

$\text{let } u = {a}^{3}$

$\Rightarrow 2 {a}^{6} - 19 {a}^{3} + 24$

$= 2 {u}^{2} - 19 u + 24$

$\text{the factors of 48 which sum to - 19 are - 3 and - 16}$

$\rightarrow 2 {u}^{2} - 16 u - 3 u + 24 \leftarrow \textcolor{b l u e}{\text{ split middle term}}$

$\text{factorise by 'grouping'}$

$= \textcolor{red}{2 u} \left(u - 8\right) \textcolor{red}{- 3} \left(u - 8\right)$

$\text{factor out } \left(u - 8\right)$

$= \left(u - 8\right) \left(\textcolor{red}{2 u - 3}\right) \leftarrow \text{ change u back to a}$

$= \left({a}^{3} - 8\right) \left(2 {a}^{3} - 3\right)$

$\left({a}^{3} - 8\right) \text{ is a "color(blue)"difference of cubes}$

â€¢color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)#

${a}^{3} - 8 = {a}^{3} - {2}^{3} = \left(a - 2\right) \left({a}^{2} + 2 a + 4\right)$

$\Rightarrow 2 {a}^{6} - 19 {a}^{3} + 24 = \left(a - 2\right) \left({a}^{2} + 2 a + 4\right) \left(2 {a}^{3} - 3\right)$