Question

What is the completely factored form of d4 -8d2 + 16

Answer 1

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 $x=d^2$:

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