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Answer 1 · 2018-03-04T18:21:23

See below.

Expanding

#a(x+r)^2+s = a(x^2+2rx+r^2)+s = a x^2+2arx+a r^2+s#

and now comparing coeficients

#{(a=3),(2ar=-18),(ar^2+s=-1):}#

and solving we get

#a = 3#
#r = -18/(2 xx 3) = -3#
#s = -1-ar^2 = -1-27 = -28#

so the answer is A

Answer 2 · 2018-03-04T18:21:34

#A#

First, we pay attention to #3x^2-18x#

We see that the coefficients share #3# as their GCF.

We can therefore rewrite this as:

#3(x^2-6x)-1#

Now, we look at the coefficient of #-6x#

The coefficient is #-6#

We now divide the coefficient by two, then square the result.

#=>(-6/2)^2=9#

#3(x^2-6x)-1# add #9# inside the parenthesis.

#3(x^2-6x+9)-1# Since we added #9# in the parenthesis, we need to make an adjustment.

We do so by multiplying #9# by the number in front of parenthesis, then subtracting it from the entire expression.

#=>3(x^2-6x+9)-1-27#

#=>3(x^2-6x+9)-28#

Factor the expression in the parenthesis.

Therefore, we now have:

#=>3(x-3)^2-28#

The answer to this problem is #-28# or #A#