Question

What is the sum of the roots of the equation
(x^2-4x+4)(x+2)(x-8) = 0

I thought it was 8 but the teacher said it's wrong

Answer 1

Discriminant of equation x² - 4 x + 4 is ∆ = ( - 4 )² - 4 ∙ 1 ∙ 4 = 16 - 16 = 0

If the discriminant is equals zero, the roots are equal.

That is called a double root.

Root of x² - 4 x + 4 = 0 is 2 multiplicity two

Root of x + 2 = 0 is - 2

Root of x - 8 = 0 is 8

So the sum of the roots = 2 + 2 - 2 + 8 = 10

Answer 2

Or, you can expand it out and use the fact that the sum of the roots of
x^4 + ax^3 + bx^2 + cx + d is -a
(x^2-4x+4)(x+2)(x-8) = x^4 - 10x^3 + 12x^2 + 40x - 64
so the sum of the roots is 10

Answer 3

or, just factor the whole thing, the last part is already done.
(x^2-4x+4)(x+2)(x-8) = 0
(x-2)(x-2)(x+2)(x-8)
the roots are, 2, 2, -2, 8
that sums to 10