Question #27939

3 Answers
Aug 2, 2015

As Sudip Sinha has pointed out -1+sqrt3i is NOT a zero. (I neglected to check that.) The other zeros are 1-sqrt3 i and 1.

Explanation:

Because all of the coefficients are real numbers, any imaginary zeros must occur in conjugate pairs.
Therefore, 1-sqrt3 i is a zero.

If c is a zero then z-c is a factor, so we could multiply

(z-(1+sqrt3 i))(z-(1-sqrt3 i)) to get z^2-2z+4
and then divide P(z) by that quadratic.

But it's quicker to consider the possible rational zero for P first. Or add the coefficients to see that 1 is also a zero.

Aug 2, 2015

1 and 1 - sqrt3 i

Explanation:

There is an error in your question. The root should be 1 + sqrt3 i . You can verify this by putting the value in the expression. If it is a root the expression should evaluate to zero.

The expression has all real coefficients, so by the Complex Conjugate Roots Theorem (https://en.wikipedia.org/wiki/Complex_conjugate_root_theorem), we have that the other complex root is 1 - sqrt3 i ,

Clearly, the third root (say a) has to be real, since it cannot have a complex conjugate; otherwise there will be 4 roots, which is not possible for a 3rd degree equation.

Note
( z - ( 1 - sqrt3 i ) ) ( z - ( 1 + sqrt3 i ) )
= ( (z - 1) + sqrt3 i ) ( (z - 1) - sqrt3 i )
= ( (z - 1)^2 - (sqrt3 i)^2 ) ( Since (z + a) (z - a) = z^2 - a^2 .)
= z^2 - 2z + 1 - 3(-1)
= z^2 - 2z + 4

We will try to get this factor in the expression.
We may write:
P(z) = z^3 - 3z^2 + 6z - 4
= z ( z^2 - 2z + 4 ) - 1 ( z^2 - 2z + 4 )
= (z - 1) ( z^2 - 2z + 4 )
= (z - 1) ( z - ( 1 - sqrt3 i ) ) ( z - ( 1 + sqrt3 i ) )

Aug 2, 2015

As an intro, I think that the root should be color(blue)(1+sqrt3) and not color(red)(-1+sqrt3)

On that basis my answer is :

z in {1," "1+sqrt3," "1-sqrt3}

Explanation:

By using the idea of complex conjugates and some other cool tricks.

P(z) is a polynomial of degree 3. This implies that it should only have 3 roots.

One interesting fact about complex roots is that they never occur alone.They always occur in conjugate pairs.

So if 1+isqrt3 is one root, then its conjugate : 1-isqrt3 most certainly is a root too!

And since there is just one more root left, we can call that root z=a.
It is not a complex number because, complex roots always occur in pairs.
And since this is the last of the 3 roots, there cannot be any other pair after the first one!

In the end the factors of P(z) were easily found to be [z-(1+isqrt3)] " , " [z-(1-isqrt3)] " and " (z-a)

NB : Note that the difference between a root and a factor is that :
- A root could be z=1+i
But the corresponding factor would be z-(1+i)

The second trick is that, by factoring P(z) we should get something like this :

P(z)=[z-(1+isqrt3)][z-(1-isqrt3)] (z-a)

Next, expand the braces,

P(z)=[z^2-z(1+isqrt3+1-isqrt3)+(1+isqrt3)(1-isqrt3)] (z-a)

=[z^2-z(2)+(1+3)] (z-a)

=[z^2-2z+4] (z-a)

=z^3+z^2(-a-2)+z(2a+4)-4a

Next, we equate this to the original polynomial P(z)=z^3-3z^2+6z-4

=>z^3+z^2(-a+2)+z(-2a+4)-4a=z^3-3z^2+6z-4

Since the two polynomials are identical, we equate the coefficients of z^3, z^2, z^1and z^0(the constant term) on either side,

Actually,we just need to pick one equation and to solve it for a

Equating the constant terms,

=>-4a=-4

=>a=1

Hence the last root is color(blue)(z=1)