factor completely:
1. 22y^4 - 33y^3 -11y^2
2. 77r^7s^7 -84r^8s^4
2007-12-10 04:17:46 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
OK
22y^4 - 33y^3 -11y^2
11y^2(2y^2 - 3y -1)
=================
77r^7s^7 -84r^8s^4
7r^7s^4(11s^3 - 12r)
==========================
Hope that helps.
2007-12-10 04:27:01 · answer #1 · answered by pyz01 7 · 0⤊ 0⤋
1. 22y^4 - 33y^3 -11y^2
= 11y^2(2y^2 -3y-1)
2. 77r^7s^7 -84r^8s^4
=7r^7s^4(11s^3- 12r^7)
2007-12-10 12:24:16 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
1)
22y^4 - 33y^3 - 11y^2
11y^2 (2y^2 -3y - 1)
2)
77(r^7)(s^7) - 84(r^8)(s^4)
7(r^7)(s^4)(11s^3-12r)
2007-12-10 12:28:18 · answer #3 · answered by xSteviex 2 · 0⤊ 0⤋
y^2(22y^2 - 33y -11)
(11y^2) * (2y^2 - 3y -1)
(11y^2) * (x -.75 -.5sqrt(3)) * (x - .75 +.5sqrt(3))
that is ugly , so I am not even going to the second one
2007-12-10 12:28:18 · answer #4 · answered by KEYNARDO 5 · 0⤊ 0⤋
1. 11y^2(2y^2-3y-1)
2. 7r^7s^4(11s^3-12r)
2007-12-10 12:26:35 · answer #5 · answered by boxer 1 · 0⤊ 0⤋