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)= (8\pm √(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`