# Decide about roots of equation (1   x   x^2) (x   x^2   1) = 0 (x^2   1   x) a) all roots are integer b) all roots are real c) all roots are rational d) all roots are double e) all roots...

(1   x   x^2)

(x   x^2   1) = 0

(x^2   1   x)

a) all roots are integer

b) all roots are real

c) all roots are rational

d) all roots are double

e) all roots aren't real

f) sum of roots is 1

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the given determinant, hence, you may add the columns 2 and 3 to the column 1, such that:

`[(1+x+x^2,x,x^2),(1+x+x^2,x^2,1),(1+x+x^2,1,x)] = 0`

Factoring out `(1+x+x^2)` yields:

`(1+x+x^2)[(1,x,x^2),(1,x^2,1),(1,1,x)]` = 0

`(1+x+x^2)(x^3 + x^2 + x - x^4 - 1 - x^2) = 0`

Reducing duplicate members in the second factor, yields:

`(1+x+x^2)(x^3 + x - x^4 - 1) = 0`

You need to group the terms in the second factor, such that:

`(1+x+x^2)((x^3 - x^4) - (1 - x)) = 0`

Factoring `x^3` in the first group yields:

`(1+x+x^2)(x^3(1 - x) - (1 - x)) = 0`

Factoring out `(1 - x)` yields:

`(1+x+x^2)(1 - x)(x^3 - 1) = 0`

Converting the difference of cubes into a product yields:

`(1 + x + x^2)(1 - x)(x - 1)(x^2 + x + 1) = 0`

`-(x^2 + x + 1)^2(x - 1)^2 = 0`

Using zero product property yields:

`(x^2 + x + 1)^2 = 0 or (x - 1)^2 = 0`

You should notice that the roots to equation `x^2 + x + 1 = 0` are complex conjugate and the square of `x^2 + x + 1` indicates the order of multiplicity of roots of equation `x^2 + x + 1 = 0` .

The equation `(x - 1)^2 = 0` has the double real root `x = 1` .

Hence, both equations `(x^2 + x + 1)^2 = 0` and `(x - 1)^2 = 0` have double roots, thus, you need to select the answer d) all roots are double.

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