Given: #(7w-14)(w^2-4w-5)=0#
We have here a three term expression that is partially broken down into its factors. We know there are three terms because the exponent or power of #w# when multiplied out will be three as in #w^3#.
Three terms need three brackets, each with a #w#. We also see there is no multiplier of #w^2#, so none are needed in the brackets:
#(7w-14)(w^2-4w-5)=0#
#(7w-14)(w ...)(w ...)=0#
Now the factors of #5# we need are #5 and 1#.
#(7w-14)(w ..5)(w ..1)=0#
The original #-5# tells us one of the factors needs to be #-#, in this case the larger factor #5# so we can obtain the middle result of #-4#.
#(7w-14)(w-5)(w+1)=0#
Then we can solve for each of the three terms:
#(7w-14)=0#
#7w=14#
#w=2#
#(w-5)=0#
#w=5#
#(w+1)=0#
#w=-1#
To check, insert values of #w# into the #given# equation. We used #-1#:
#(7w-14)(w^2-4w-5)=0#
#(7(-1)-14)((-1)^2-4(-1)w-5)=0#
#(-7-14)(1+4-5)=0#
#(-21)(0)=0#
#0=0#
