(4-m)x^2 + (2m+4) x + (8m+1)

For the perect square , we know that the discrminant should equal zero.

==> delta = b^2 - 4ac = 0

==> (2m + 4)^2 - 4(4-m)(8m+1) = 0

Let us expand brackets:

==> 4m^2 + 16m + 16 + 32m^2 - 124 m -16 = 0

==> 36m^2 - 108m = 0

==> 36m(m - 3) = 0

==> there is two solutions:

m1= 0

m2= 3

The expression (4-m)*x^2+(2m+4)*x+(8m+1) will be a perfect square, if the equation (4-m)*x^2+(2m+4)*x+(8m+1) = 0 will have 2 equal roots.

For the equation to have equal roots, the discriminant of the equation, delta, will have to be equal to zero.

delta = b^2 - 4*a*c, where a,b,c are the coefficients of the equation.

Let's identify a,b,c:

a = 4-m

b = 2m+4

c = 8m + 1

delta = (2m+4)^2 - 4*(4-m)(8m+1)

delta = 0 => (2m+4)^2 - 4*(4-m)(8m+1) = 0

We'll remove the brackets:

4m^2 + 16m + 16 - 4(32m+4-8m^2-m) = 0

4m^2 + 16m + 16 + 32m^2 - 124m - 16 = 0

We'll combine and eliminate like terms:

36m^2 - 108m = 0

We'll factorize by 36m:

36m(m-3)=0

We'll set each factor as 0.

36m = 0

m = 0

m-3 = 0

m = 3

**So, the expression is a perfect square for m = {0 ; 3}.**

To find values of m for which (4-m)x^2+(2m+4)x+(8m+1) is a perfect square.

We know that (ax+b)^2 = a^2x^2+2abx+b^2 is a perfect square. The condtion is Product of 4 times coefficient of x^2 and constant term = square of coefficient of x.

So in Ax^2+Bx+C, expression becomes perfect square if 4AC = B^2.

So here, A= a-m, B = 2m+2 and C = 8m+1

4(4-m)(8m+1) = (2m+4)^2.

4(32m+4-8m^2-m) = 4m^2+16m+16

-32m^2+128m-4m +16 = 4m^2+16m+16

0 = 36m^2+(16-128+4)m +16-16

0= 36m^2 -108m

0 = 36m(m-3)

m = 0 or m =3.

Verification:

m =0: (4-m)x^2+(2m+4)x+(8m+1) = 4x^2+4x+1 = (2x+1)^2

m=3: (4-3)x^2 +(2*3+4)+(8*3+1) = x^2+5x+25 = (x+5)^2.