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)#