2007-05-21 07:17:06 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
y^2 +9y-4y-36=0
y(y+9)-4(y+9)=0
(y+9)(y-4)=0
2007-05-21 07:21:19 · answer #1 · answered by Maths Rocks 4 · 1⤊ 0⤋
Factor: y^2 + 5y - 36
(y + 9)(y - 4)
.
2007-05-21 07:21:33 · answer #2 · answered by Robert L 7 · 1⤊ 0⤋
Find all multiples of negative 36. then find the sum that adds to positive 5.
There is
-36 and 1
36 and -1
-18 and 2
18 and -2
-12 and 3
12 and -3
-9 and 4
9 and -4
To get positive 5 use 9 and -4.
Therefore the answer is (y + 9)(y - 4)
2007-05-21 07:22:09 · answer #3 · answered by tntigers99 1 · 1⤊ 0⤋
y^2+5y-36
y^2+9y-4y-36 [Since 9*4=36 and 9-4=5]
y(y+9)-4(y-9)=0
(y-4)(y+9)=0
So y=4 or -9
2007-05-21 07:32:09 · answer #4 · answered by godisgreat 1 · 1⤊ 0⤋
36=6*6 or 9*4
use 9*4 because difference is 5
(y+9)(y-4)
2007-05-21 07:46:31 · answer #5 · answered by jon d 3 · 1⤊ 0⤋
y^2+5y-36=(y-y')(y-y'') in factor.
solve for y' and y'':
det=b^2-4ac
with :b=5 ,a=1 and c=-36
det=5^2-4(1)(-36)
det=25+144
det=169
sqrt (det)=sqrt169=13
then,
y=[-b+or-sqrt(det)]/2a
y=[-5+or-13]/2
y'=(-5+13)/2 and y''=(-5-13)/2
y'=4 and y'=-9
then,
(y-y')(y-y'')=(y-4)(y+9)
2007-05-21 07:47:19 · answer #6 · answered by Johnny 2 · 1⤊ 0⤋
(y+9)(y-4)=0
2007-05-21 07:20:52 · answer #7 · answered by Yssa A 3 · 1⤊ 0⤋
= (y + 9).(y - 4)
2007-05-21 10:33:11 · answer #8 · answered by Como 7 · 1⤊ 0⤋
(y+9)*(y-4)
2007-05-21 07:20:04 · answer #9 · answered by Anonymous · 1⤊ 0⤋