How to factorise 2a6−19a3+24?
2 Answers
Oct 5, 2017
Explanation:
let u=a^3#
we have then
a quadratic in
factors of
we have
substituting back
the second bracket is difference of cubes
Oct 5, 2017
Explanation:
let u=a3
⇒2a6−19a3+24
=2u2−19u+24
the factors of 48 which sum to - 19 are - 3 and - 16
→2u2−16u−3u+24← split middle term
factorise by 'grouping'
=2u(u−8)−3(u−8)
factor out (u−8)
=(u−8)(2u−3)← change u back to a
=(a3−8)(2a3−3)
(a3−8) is a difference of cubes
∙xa3−b3=(a−b)(a2+ab+b2)
a3−8=a3−23=(a−2)(a2+2a+4)
⇒2a6−19a3+24=(a−2)(a2+2a+4)(2a3−3)