2007-05-09 13:14:53 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
5y^3+10y^2-25y
5y(y^2+2y-5)
2007-05-09 14:19:49 · answer #1 · answered by bootis32 6 · 0⤊ 0⤋
Basically you have trhee groups of numbers:
1) 5y^3
2) 10y^2
3) 25y
You have to pick put the smallest thing they all have in common (what could go into them evenly). 5 can go into them all evenly and y can as well. Therefore, you can factor out 5y. You are left with 5y(y^2 + 2y + 5)
2007-05-09 20:21:22 · answer #2 · answered by Carolyn D 5 · 0⤊ 0⤋
5y is common in each term therefore the equation becomes
5y(y^2 + 2y - 5)
2007-05-09 20:28:40 · answer #3 · answered by bill t 2 · 0⤊ 0⤋
5y (y^2 + 2y - 5) ?
2007-05-09 20:20:33 · answer #4 · answered by Anonymous · 0⤊ 0⤋
5y.(y² + 2y - 5) is as far as it goes.
2007-05-10 09:22:28 · answer #5 · answered by Como 7 · 0⤊ 0⤋