How do you factor 10n-(2n-3)^210n(2n3)2?

1 Answer
Mar 11, 2018

First expand.
10n - 4n^2 -6n + 910n4n26n+9
= -4n^2 + 4n + 9=4n2+4n+9
Factorise out the -1.
= -1(4n^2 -4n -9)=1(4n24n9)
Now use quadratic equation.
a = 4, b = -4, c = -9.
(4 +- sqrt(16+144))/-724±16+14472
Then you are left with (when simplified):
(1 +- sqrt(10))/181±1018
Now put into factorised form...
(18n +1 - sqrt(10))(n + 1 + sqrt(10))(18n+110)(n+1+10)