Perfect Square problem

If 9x^4-12x^3+28x^2+ax+b is a perfect square, find the value of a and b.
This is one of my past year examination's questions, some help on it?

Answer:
a = - 16 and b = 16 worked out as under.

Let 9x^4 - 12x^3 + 28x^2 + ax + b
= (3x^2 - cx + d)^2
= 9x^4 - 6cx^3 + (6d + c^2)x^2 + (-2cd)x + d^2

=> -6c = -12 => c = 2
6d + c^2 = 28
=> 6d = 28 - c^2 = 24
=> d = 4
a = - 2cd = - 2*2*4 = - 16 and
b = d^2 = 4^2 = 16
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Verification:
9x^4 - 12x^3 + 28x^2 - 16x + 16
= (3x^2 - 2x + 4)^2

(cx² + dx + e)² = (cx² + dx + e)(cx² + dx + e)
(cx² + dx + e)² = c²x⁴ + 2cdx³ + 2cex² + d²x² + 2dex + e²
(cx² + dx + e)² = c²x⁴ + 2cdx³ + (2ce + d²)x² + 2dex + e²

c²x⁴ + 2cdx³ + (2ce + d²)x² + 2dex + e² = 9x⁴ - 12x³ + 28x² + ax + b

Looking at the x⁴ term, c² = 9 → c = ±3
Looking at the x³ term, 2cd = -12 → d = ∓2
Looking at the x² term, (2ce + d²) = 28 → e = ±4

Therefore:
a = 2de
a = -2(2)(4) = -16

b = e²
b = 16

suppose it is a perfect square (sx^2 + tx + u)^2 =
s^2x^4 + 2stx^3 + (2su + t^2)x^2 + 2tux + u^2
so s^2 = 9 (s = +-3)
2st = -12 (t = -+2
2su + t^2 = 28, so 2su = 24, u = +-4
therefore b = 16
a = 2tu = -16 (2 * 2 * -4 or 2 * -2 * 4)

the perfect square:
(kx^2+mx+n)^2
breaking it down...
k^2*x^4 + (2km)x^3 + (2kn+m^2)x^2 + (2mn)x + n^2

so
9 = k^2
k = 3 or -3

-12 = 2km
m = -2 or 2

28 = 2kn + m^2
n = 7/3 or -7/3

a = 2mn = -14/3

b = n^2 = 49/9

(x + a)^4 = (x + a)^2 (x + a)^2
= (x^2 + a^2 + 2ax) (x^2 + a^2 + 2ax)
= x^4 + x^2 a^2 + 2ax^3 + x^2a^2 + a^4 + 2a^3x + 2ax^3 + 2a^3x + 4a^2 x^2
= x^4 + a^4 + 6a^2x^2 + 4ax^3 + 4a^3x

compare 9x^4-12x^3+28x^2+ax+b
4a = -12 , a = -3

thus b =a^4 = 81
a = -3

(3x^2-2x+4)^2=9x^4-12x^3+28x^2_16x+16 , a=-16, b=16