Respuesta :

According to Vieta's Formulas, if [tex]x_1,x_2[/tex] are solutions of a given quadratic equation:

[tex]ax^2+bx+c=0[/tex]

Then:

[tex]a(x-x_1)(x-x_2)[/tex] is the completely factored form of [tex]ax^2+bx+c[/tex].

If choose [tex]x=d^2[/tex], then:

[tex]\displaystyle x^2-8x+16=0\\\\x_{1,2}= \frac{8\pm \sqrt{64-64} }{2}=4 [/tex]

So, according to Vieta's formula, we can get:

[tex]x^2-8x+16=(x-4)(x-4)= (x-4)^2[/tex]

But [tex]x=d^2[/tex]:

[tex]d^4-8d^2+16=(d^2-4)^2=[(d+2)(d-2)]^2=(d+2)^2(d-2)^2[/tex]

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