How do you solve #-2( w + 1) ( - w - 4) = 0#?

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Answer 1 · 2018-01-23T05:37:50

We have two solutions for #w#:

#w=-1#

#w=-4#

Well, the question says

#-2(w+1)(-w-4)=0#

So, dividing by #-2# for both sides gives us

#(w+1)(-w-4)=0/-2#

#(w+1)(-w-4)=0#

So now, we have two cases:

#w+1=0#

#-w-4=0#

Solving for both cases, we get:

#w=-1#

#w=-4#

That means, that there are two solutions for #w#:

#w=-1,-4#