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

2 Answers
sjc
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)#

Jim G.
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)#

Related questions
Impact of this question
1586 views around the world
You can reuse this answer
Creative Commons License