3y^2 - 11y - 20
2007-07-25 07:01:41 · 9 answers · asked by chrizzle08 2 in Science & Mathematics ➔ Mathematics
(3y + 4)(y - 5)
this is a tough one. the best way to do this is break 20 up into its multiples:
1 x 20
2 x 10
4 x 5
then figure out what combinations works..
2007-07-25 07:04:12 · answer #1 · answered by BrightEyes 5 · 1⤊ 0⤋
3y^2 -11y -20=0
(3y +4)(y- 5)= 3y^2 +4y- 15y -20= 0
3y+4=0 so 3y= -4 so y= -4/3
& y-5=0 so y=5
2007-07-25 07:13:13 · answer #2 · answered by Just me 5 · 0⤊ 0⤋
3y²-11y-20
=3y² - 15y + 4y - 20
= 3y(y-5) +4(y-5)
= (3y+4)(y-5) . . . . . Ans
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2007-07-25 07:10:01 · answer #3 · answered by Joymash 6 · 0⤊ 0⤋
(3y + 4) (y - 5)
Check
3y (y - 5) + 4(y - 5)
3y² - 15y + 4y - 20
3y² - 11y - 20 (as required)
2007-07-25 07:37:16 · answer #4 · answered by Como 7 · 0⤊ 0⤋
3y^2-11y-20
3y^2+4y-15y-20
or y(3y+4)-5(3y+4)
or (y-5)(3y+4) ans
2007-07-25 07:57:23 · answer #5 · answered by MAHAANIM07 4 · 0⤊ 0⤋
3y^2-11y-20=3y^2-15y+4y-20
=3y(y-5)+4(y-5)
=(y-5)(3y+4).ANS
2007-07-25 07:07:56 · answer #6 · answered by Anonymous · 1⤊ 0⤋
(3y - 4) (y + 5)
2007-07-25 07:05:27 · answer #7 · answered by adambauman31 2 · 1⤊ 0⤋
let me fink...
(3y - 4) (y + 5)
yup luks ryt :D x
2007-07-25 07:07:06 · answer #8 · answered by Anonymous · 1⤊ 0⤋
(y-3)(y-4/3)
I think.
2007-07-25 07:09:49 · answer #9 · answered by blondeegirl 2 · 0⤊ 1⤋