Question 890102
I will do just number 1.  You can then understand the process and do number 2 yourself.


{{{y=3(x^2-3x-28)}}}  multiplied the binomials but not the separate 3 factor.
The term which completes the square is {{{(3/2)^2}}}.
{{{y=3(x^2-3x+(3/2)^2-(3/2)^2-28)}}}--add and subtract the square term inside parentheses
{{{y=3((x-3/2)^2-9/4-28)}}}---factorize the square trinomial part
{{{3((x-3/2)^2-9/4-112/4)}}}
{{{3((x-3/2)^2-116/4)}}}
{{{3((x-3/2)^2-29)}}}
{{{y=3(x-3/2)^2-3*29}}}
{{{highlight(y=3(x-3/2)^2-87)}}}




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See a separate lesson on Completing The Square.