(3y-2)^2-(y+1)^2
Your help is greatly appreciated:)
2007-05-05 18:14:38 · 4 answers · asked by tickled pink 1 in Science & Mathematics ➔ Mathematics
(3y-2)^2-(y+1)^2
= 9y^2 -12y + 4
-(y^2 + 2y +1)
= 8y^2 -14y + 3
here I use the quadratic formula:
y = (-b PORM sqrt(b^2 - 4ac) )/ 2a
[porm = "PLUS OR MINUS"]
y= (14 porm (sqrt(196 - 96)) / 16
y = (14 porm 10) / 16
y = 24/16 or 4/16 = 3/2 or 1/4
so the final factorizatyion would be
(y - 3/2)*(y - 1/4)* k [where "k" is a constant ... make the fractions disappear by multiplying thru by "8"
(2y - 3)* (4y -1) <==== "answer"
check:
(2y - 3)* (4y -1) = 8y^2 -12y -2y +3 = 8y^2 -14y +3
looks right
2007-05-05 18:17:49 · answer #1 · answered by atheistforthebirthofjesus 6 · 0⤊ 0⤋
It is simple We have a^2-b^2=(a-b)(a+b) so if we take a=3y-2 and b=y+1 we have
a^2-b^2=(a-b)(a+b)=(3y-2-y-1)(3y-2+y+1)=(2y-3)(4y-1)
2007-05-06 01:27:59 · answer #2 · answered by Ahmad k 2 · 0⤊ 0⤋
Use the fact that a^2-b^2=(a+b)(a-b). This will make your factoring problem easy.
2007-05-06 01:19:31 · answer #3 · answered by Scott H 3 · 0⤊ 0⤋
(3y-2)^2-(y+1)^2
=(3y-2-y-1)(3y-2+y+1)
=(2y-3)(4y-1)
2007-05-06 01:22:31 · answer #4 · answered by iyiogrenci 6 · 0⤊ 0⤋