# complex square rootif polynomial is p=x^4-4x^2+16, x=square root 3 -i may be a root?

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### 2 Answers

To test if 3 - i is a root of p=x^4-4x^2+16, substitute x = sqrt 3 - i

x^4 - 4x^2 + 16

=>(sqrt 3- i)^4 -4(sqrt 3- i)^2 + 16

=> (sqrt 3 - i)^2^2 - 4*(3 + i^2 - 2*i*sqrt 3) + 16

=> (3 + i^2 - 2*sqrt 3*i)^2 - 12 + 4 + 8*sqrt 3*i + 16

=> 4 + 4*3*i^2 - 8sqrt 3*i - 8 + 8*sqrt 3*i + 16

=> 4 - 12 - 8sqrt 3*i - 8 + 8*sqrt 3*i + 16

=> 0

**As substituting x = sqrt 3 - i gives the value of the polynomial as 0, it is a root.**

We'll substitute x by sqrt3 - i and we'll verify if f(sqrt3 - i) = 0.

f(x) = x^4 – 4x^2 + 16

f(x) = x^2(x^2 - 4) + 16

We'll re-write the difference of squares x^2 - 4 = (x-2)(x+2)

x = sqrt3 - i

We'll square raise both sides:

x^2 = (sqrt3 - i)^2

We'll expand the square:

x^2 = 3 - 2isqrt3 + i^2, where i^2 = -1

x^2 = 2 - 2isqrt3

f(sqrt3 - i) = (2 - 2isqrt3)(2 - 2isqrt3 - 4) + 16

We'll combine like terms inside brackets:

f(sqrt3 - i) = (2 - 2isqrt3)(-2 - 2isqrt3) + 16

f(sqrt3 - i) = -(2 - 2isqrt3)(2 + 2isqrt3) + 16

We'll write the product as a difference of squares:

f(sqrt3 - i) = -(2^2 - 4*3*i^2) + 16

f(sqrt3 - i) =- (4 + 12) + 16

f(sqrt3 - i) = -16 + 16

f(sqrt3 - i) = 0 => sqrt3 - i is the root of the polynomial