16y^2(8y)=0?
Factor out an 8y to give you 8y(2y(2))=0
Use order of operations to give you 8y(4y)=0
Distribute to give you 32y=0
and ultimately the answer will be 0
If there is a missing plus or minus operator between 16y2 and the 8y, then it can be factorised by taking out common factors:
16y2 ± 8y = 0
⇒ 8y(2y ± 1) = 0
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y2 +8y + 16 = 0 can factor to (y+4) (y+4) = 0 so y+4 = 0 so y = -4
x = 2, y = 84x + 8y= 4(2) + 8(8)= 8 + 64= 72
Not sure which you mean, take your pick: (8y)3+27=(8y+3)((8y)2-3*(8y)+9)=(8y+3)(64y2-24y+9); or 8y3+27=(2y+3)(4y2-6y+9)
8y + 7y = 15ySo [ 8y + 9 + 7y ] = 15y + 9If you want to get fancy and factor that expression, it's 3(5y + 3) .
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