Factor completely the Polynomial below: 2a^2-4a-2?

Answer 1

if you know the quadratic equation of the general form (ax^2+bx+c=0) .... it has real roots (and hence real factors) only if the discriminant (b^2-4ac) is > 0

here a=2, b= -4, c= -2

so discriminant = 16 + 4*2*2 = 32 > 0; so it has real factors

roots are given by (4 +/- sqrt (32))/4 or 1 +/- sqrt (2)

so the factors are 2 * (a - 1 - sqrt(2)) * (a - 1 + sqrt(2))

this can also be done as: 2 (a^2 - 2a +1 - 2) = 2 ((a-1)^2 - sqrt(2)^2)

using the algebraic identity to factorise now...

Answer 2

First, you can factor out a 2

2(a^2 - 2a - 1)

Are you sure the signs of the trinomial are correct?

a^2 - 2a + 1 = (a - 1)(a - 1) and

a^2 + 2a + 1 = (a + 1)(a + 1) but

a^2 - 2a - 1 = ????