Question 82235
1.factor: y2-4y-5
(y-5)(y+1)
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2. factor 3a2+9a+6
(3a+3)(a+2)
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3.evaluate: x(x+1)(x-1)
set each factor equal to zero and solve for the x-term:
x=0
x+1=0
x=-1
and x-1=0
x=1
So, x= 0, -1 and 1
.

4.factor: 6y2-2yz
factor out (2y)
{{{(6y^2/2y-2yz/2y)}}}
(2y)(3y-z)
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