Question 268873
find c so that the polynomial will be the square of a bynomial cy2+6y+1  justify your answer.
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Assume A and B are such that

{{{(Ay+B)^2=cy^2+6y+1}}}

{{{(Ay+B)(Ay+B)=cy^2+6y+1}}}

{{{A^2y^2+2ABy+B^2=cy^2+6y+1}}}

To make that an identity we equate like terms on both sides:

{{{system(A^2y^2=cy^2,2ABy=6y,B^2=1)}}}

simplify

{{{system(A^2=c,AB=3,B^2=1)}}}

Solving the last equation 

{{{B^2=1}}},

{{{B^2-1=0}}}

{{{(B-1)(B+1)=0}}}

Setting first factor = 0

{{{B-1=0}}}
{{{B=1}}}

Setting second factor = 0

{{{B+1=0}}}
{{{B=-1}}}

Substituting {{{B=1}}} in the equation

{{{AB=3}}}

{{{A*(1)=3}}}

{{{A=3}}}

Substituting that in

{{{A^2=c}}}

{{{3^2=c}}}

{{{9=c}}}

So

{{{(Ay+B)^2=cy^2+6y+1}}}

becomes

{{{(3y+1)^2=9y^2+6y+1}}}

If you use {{{B=-1}}}, you end up with

{{{(-3y-1)^2=9y^2+6y+1}}}

Either way, {{{C=9}}}

Edwin</pre>