4y²-8y-12? I think I might almost have it down pat.?

Question

I have got 2(y²-4y-6) is that right? And can you factor even lower?

Answer 1

(4y-12) (y+1)

4(y-3)(y+1)

Answer 2

4(y^2-2y-3)

4(y-3)(y+1)

Answer 3

4y²-8y-12

4(y²-2y-3)

4(y - 3)(y + 1)

Answer 4

No you aren't right...let me correct you:

4y^2 - 8y - 12

4 ( y^2 - 2y - 3 )      {taking common}

4 ( y^2 - 3y + y -3 )     {breaking the middle term}

4 [y (y - 3 ) + 1 ( y - 3) ]      {taking common}

4(y - 3) (y+1) ANSWER

But tell me one thing .... how did you wrote the power of y like that??? Help in additional information please

Answer 5

not quite:

4y²-8y-12

=4 (y^2 - 2y -3)

=4 (y-3) (y+1)

...

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and if you set it = 0

ie:

4y²-8y-12 = 0

4(y-3) (y+1) = 0

(y-3) (y+1) = 0

means

y = 3 or y= -1 are "zeros" of the equation 4y²-8y-12 = 0