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

2 Answers
Oct 5, 2017

(2a^3-3)(a-2)(a^2+2a+4)

Explanation:

let u=a^3#

we have then

2u^2-19a^3+24

a quadratic in a^3

2xx24=48

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

we have

2u^2-3u-16u+24

=(2u^2-3u)-(16u-24)

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

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

substituting back

(2a^3-3)(a^3-8)

the second bracket is difference of cubes

(2a^3-3)(a-2)(a^2+2a+4)

Oct 5, 2017

(a-2)(a^2+2a+4)(2a^3-3)

Explanation:

"let " u=a^3

rArr2a^6-19a^3+24

=2u^2-19u+24

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

rarr2u^2-16u-3u+24larrcolor(blue)" split middle term"

"factorise by 'grouping'"

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

"factor out "(u-8)

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

=(a^3-8)(2a^3-3)

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

•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+2a+4)

rArr2a^6-19a^3+24=(a-2)(a^2+2a+4)(2a^3-3)