How do you find all zeros of the function $P (x) = 2 x^{3} - 5 x^{2} + 6 x - 2$ $P(x)=2x^3-5x^2+6x-2$ given the zero 1+i?

Question

How do you find all zeros of the function $P (x) = 2 x^{3} - 5 x^{2} + 6 x - 2$ $P(x)=2x^3-5x^2+6x-2$ given the zero 1+i?

1 Answer

Impact of this question

2964 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-16T22:35:25

Zeros: $1 + i$ $1+i$ , $1 - i$ $1-i$ , $\frac{1}{2}$ $1/2$

Explanation:

$P (x) = 2 x^{3} - 5 x^{2} + 6 x - 2$ $P(x) = 2x^3-5x^2+6x-2$

We are told that $x = 1 + i$ $x=1+i$ is a zero.

Since the coefficients of $P (x)$ $P(x)$ are all Real, the Complex conjugate $1 - i$ $1-i$ must be a zero too. So $P (x)$ $P(x)$ must be divisible by:

$(x - (1 - i)) (x - (1 + i))$ $(x-(1-i))(x-(1+i))$

$= ((x - 1) - i) ((x - 1) + i)$ $= ((x-1)-i)((x-1)+i)$

$= {(x - 1)}^{2} - i^{2}$ $= (x-1)^2-i^2$

$= x^{2} - 2 x + 2$ $= x^2-2x+2$

We find:

$2 x^{3} - 5 x^{2} + 6 x - 2 = (x^{2} - 2 x + 2) (2 x - 1)$ $2x^3-5x^2+6x-2 = (x^2-2x+2)(2x-1)$

So the remaining zero is $x = \frac{1}{2}$ $x=1/2$

Science

Math

Humanities

... and beyond

How do you find all zeros of the function $P (x) = 2 x^{3} - 5 x^{2} + 6 x - 2$ $P(x)=2x^3-5x^2+6x-2$ given the zero 1+i?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find all zeros of the function P(x)=2x^3-5x^2+6x-2P(x)=2x3−5x2+6x−2P(x)=2x^3-5x^2+6x-2 given the zero 1+i?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find all zeros of the function $P (x) = 2 x^{3} - 5 x^{2} + 6 x - 2$ $P(x)=2x^3-5x^2+6x-2$ given the zero 1+i?