3n2+3n-90 ?

3n² + 3n - 90
= 3(n² + n - 30)
= 3(n² - 5n + 6n - 30)
= 3(n(n-5)+6(n-5))
= 3(n+6)(n-5)

3n² + 3n - 90
3(n² + n - 30)
3(n + 6)(n - 5)

3(n^2+n-30 )====> 3(n+6)(n-5)

3 ( n ² + n - 30 )
3 ( n + 6 ) ( n - 5 )

ninety:Ice cube Amerikkka's maximum needed ninety one: A Tribe talked approximately as Quest A low end concept ninety two. Pete Rock and CL mushy Mecca and the Soul Brother ninety 3: Black Moon Entra da point ninety 4: NaS Illmatic ninety 5: GZA Liquid Swords ninety six: Ghost face Killah: Ironman ninety seven: wU tANG continuously ninety 8. Mos Def and Talib Kweli: Black celebrity ninety 9: Dr. Dre. 2001

hi!

3n^2 + 3n - 90

factor out 3 first:
3(n^2 + n - 30)

factor (n^2 + n - 30) and multiply it with 3
3(n - 6)(n + 5)

i hope this helps

=)

First divide the 3 out of each term:
=3(n^2+n-30)
Then do your usual factoring method (what two numbers multiply to -30 and add to 1), which are 6 and -5:
=3(n+6)(n-5)

PS- n^2 means n squared :)

Theres a trick for these:

Take 3 from the n^2 and the -90 and multiply to get -270.

Now you need two numbers whose product is -270 and

whose sum is equal to 3. That would be 18 and -15, since

18 + -15 = 3 and 18 x -15 = -270. Rewrite expression as:

3n^2 + 18n - 15n - 90 and factor:

3n(n + 6) -15(n + 6) = (3n - 15)(n + 6) = 3(n - 5)(n + 6).

If you can factor, this method always works.

3n^2 + 3n - 90
=3(n+6)(n-5)


how did some of my above authors got the 2 in superscript

3n^2 + 3n - 90
= 3(n^2 + n - 30)
= 3(n + 6)(n - 5)