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Solve the equation by factoring. 3z^2 + 3z - 6 =0
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Pull out the three, you get 3(z-1)(z+2)=0
Then divide everything by 3.
\[ax^2+bx+c\] to factor this we find two factors of a*c that have product a*c and have sum b a=3, b=3,c=-6 a*c=3(-6)=-3(6) b=6-3 so we have \[3x^2+6x-3x-6=0\] \[3x(x+2)-3(x+2)=0\] \[(x+2)(3x-3)=0\] \[3(x+2)(x-1)=0\]
so Z=1 or z=-2
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why have i see this an similar problem and no one divide by 3 first?
\[3 \neq 0, x+2=0, x-1=0=>x=-2 or x=1\]
where x=z lol
\[z^2+z-2=0\] is a good first step
why factor out 3? just get rid of it, after all 0/3 is still 0
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