According to Vieta's Formulas, if \(x_1,x_2\) are solutions of a given quadratic equation:
\(ax^2+bx+c=0\)
Then:
\(a(x-x_1)(x-x_2)\) is the completely factored form of \(ax^2+bx+c\).
If choose \(x=d^2\), then:
\(\displaystyle x^2-8x+16=0\\\\x_{1,2}= \frac{8\pm \sqrt{64-64} }{2}=4\)
So, according to Vieta's formula, we can get:
\(x^2-8x+16=(x-4)(x-4)= (x-4)^2\)
But \(x=d^2\):
\(d^4-8d^2+16=(d^2-4)^2=[(d+2)(d-2)]^2=(d+2)^2(d-2)^2\)