SOLUTION: 91z+30z^2-30 factored completely

Question 652348: 91z+30z^2-30 factored completely
Answer by Shana-D77(132) (Show Source): You can put this solution on YOUR website!
It's best to rearrange this to 30z^2 + 91z - 30 (Az^2 + Bz + C)
First, multiply A*C (30*-30 = 900)
Ask "What two numbers multiply to get -900 that add to get B (91)?"
Answer: 100*-9 (trial and error)
Rewrite B using the pair you just found:
30z^2 + 91z - 30
30z^2 + 100z - 9z - 30
Parenthesize:
(30z^2 + 100z) - (9z - 30)
Factor each parenthesis:
10z(3z + 10) - 3(3z + 10) <<- the parenthesis will always be the same. The first one is always right.
Rewrite: One () is the stuff on the inside once, the other () is the stuff on the outside:
(3z + 10)(10z - 3)
RELATED QUESTIONS
Facto: 30z^2 - 87z +... (answered by mollukutti)

Factor... (answered by JBarnum)

Factor completely 30z^8 + 44z^5 +... (answered by MRperkins)

Factor completely. If a polynomial cannot be factored, write prime.... (answered by ewatrrr)

5w^2-45 factored out... (answered by drj)

10r^2-15r-175 factored... (answered by kijj95)

8z^4-32z^3+30z^2 (answered by jim_thompson5910)

Factor each polynomial as completely as possible...If a polynomial cannot be factored... (answered by queenofit)

Is this equation factored completely : 2(2x^2-9) (answered by Alan3354)