How to factorise $2 a^{6} - 19 a^{3} + 24$ $2a^6-19a^3+24$ ?

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How to factorise $2 a^{6} - 19 a^{3} + 24$ $2a^6-19a^3+24$ ?

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Answer 1 · 2017-10-05T12:52:32

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

Explanation:

let u=a^3#

we have then

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

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

$2 \times 24 = 48$ $2xx24=48$

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

we have

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

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

$= u (2 u - 3) - 8 (2 u - 3)$ $=u(2u-3)-8(2u-3)$

$= (2 u - 3) (u - 8)$ $=(2u-3)(u-8)$

substituting back

$(2 a^{3} - 3) (a^{3} - 8)$ $(2a^3-3)(a^3-8)$

the second bracket is difference of cubes

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

Answer 2 · 2017-10-05T13:16:34

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

Explanation:

$let u = a^{3}$ $"let " u=a^3$

$\Rightarrow 2 a^{6} - 19 a^{3} + 24$ $rArr2a^6-19a^3+24$

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

$the factors of 48 which sum to - 19 are - 3 and - 16$ $"the factors of 48 which sum to - 19 are - 3 and - 16"$

$\to 2 u^{2} - 16 u - 3 u + 24 \leftarrow split middle term$ $rarr2u^2-16u-3u+24larrcolor(blue)" split middle term"$

$factorise by 'grouping'$ $"factorise by 'grouping'"$

$= 2 u (u - 8) - 3 (u - 8)$ $=color(red)(2u)(u-8)color(red)(-3)(u-8)$

$factor out (u - 8)$ $"factor out "(u-8)$

$= (u - 8) (2 u - 3) \leftarrow change u back to a$ $=(u-8)(color(red)(2u-3))larr" change u back to a"$

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

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

$∙ x a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)$

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

$\Rightarrow 2 a^{6} - 19 a^{3} + 24 = (a - 2) (a^{2} + 2 a + 4) (2 a^{3} - 3)$ $rArr2a^6-19a^3+24=(a-2)(a^2+2a+4)(2a^3-3)$

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How to factorise $2 a^{6} - 19 a^{3} + 24$ $2a^6-19a^3+24$ ?

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