How to factorise 2a619a3+24?

2 Answers
Oct 5, 2017

(2a33)(a2)(a2+2a+4)

Explanation:

let u=a^3#

we have then

2u219a3+24

a quadratic in a3

2×24=48

factors of 48 that sum to 1916,3

we have

2u23u16u+24

=(2u23u)(16u24)

=u(2u3)8(2u3)

=(2u3)(u8)

substituting back

(2a33)(a38)

the second bracket is difference of cubes

(2a33)(a2)(a2+2a+4)

Oct 5, 2017

(a2)(a2+2a+4)(2a33)

Explanation:

let u=a3

2a619a3+24

=2u219u+24

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

2u216u3u+24 split middle term

factorise by 'grouping'

=2u(u8)3(u8)

factor out (u8)

=(u8)(2u3) change u back to a

=(a38)(2a33)

(a38) is a difference of cubes

xa3b3=(ab)(a2+ab+b2)

a38=a323=(a2)(a2+2a+4)

2a619a3+24=(a2)(a2+2a+4)(2a33)