How to factorise 2a^6-19a^3+24?
2 Answers
Explanation:
let u=a^3#
we have then
a quadratic in
factors of
we have
substituting back
the second bracket is difference of cubes
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)