Faktorisasi Suku Aljabar 2.(-2y-3)²=... a.4y²+6y+9 b.-4y²+6y+9 c.4y²-6y+9 d.-4y²+6y-9 4.-6p²+16p-8=… a.(3p+2) (2p-4) b.(-3p+2) (2p-4) c(-3p+2) (-2p-4) d.(-3p+2)

Nomor 2
= (-2y - 3)²
= (-2y - 3)(-2y - 3)
= 4y² + 6y + 6y + 9
= 4y² + 12y + 9

Opsi gak ada

Nomor 3
-6p² + 16p - 8

a = -6
b = 16
c = -8

X1,2
= -b ± √(b² - 4ac) / 2a
= -16 ± √(16² - 4.(-6).(-8) / 2(-6)
= -16 ± √(256 - 192) / -12
= (-16 ± √64) / -12
= (-16 ± 8) / -12

X1 = (-16 + 8) / -12
X1 = -8 / -12
X1 = 2/3

X2 = (-16 - 8) / -12
X2 = -24 / -12
X2 = 2

maka
-6p² + 16p - 8 → (3p - 2)(2p - 4)
....................... → (-3p + 2)(-2p + 4)

Opsi D

semoga berguna +_+