How do you find all the roots of $f (x) = x^{4} + 2 x^{3} + x^{2} - 2 x - 2$ $f(x) = x^4 + 2x^3 + x^2 - 2x - 2$ ?

Question

How do you find all the roots of $f (x) = x^{4} + 2 x^{3} + x^{2} - 2 x - 2$ $f(x) = x^4 + 2x^3 + x^2 - 2x - 2$ ?

1 Answer

Impact of this question

2059 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-23T18:20:06

This quartic polynomial has zeros $\pm 1$ $+-1$ and $- 1 \pm i$ $-1+-i$ .

Explanation:

First note that the sum of the coefficients is $0$ $0$ . That is, $1 + 2 + 1 - 2 - 2 = 0$ $1+2+1-2-2 = 0$ . Hence we can deduce that $x = 1$ $x=1$ is a zero and $(x - 1)$ $(x-1)$ a factor:

$x^{4} + 2 x^{3} + x^{2} - 2 x - 2 = (x - 1) (x^{3} + 3 x^{2} + 4 x + 2)$ $x^4+2x^3+x^2-2x-2 = (x-1)(x^3+3x^2+4x+2)$

If you reverse the signs of the coefficients of the terms of odd degree in the remaining cubic $(x^{3} + 3 x^{2} + 4 x + 2)$ $(x^3+3x^2+4x+2)$ you get: $- 1 + 3 - 4 + 2 = 0$ $-1+3-4+2 = 0$ . Hence $x = - 1$ $x=-1$ is a zero and $(x + 1)$ $(x+1)$ a factor:

$x^{3} + 3 x^{2} + 4 x + 2 = (x + 1) (x^{2} + 2 x + 2)$ $x^3+3x^2+4x+2 = (x+1)(x^2+2x+2)$

The remaining quadratic factor has negative discriminant, but you can factor it by completing the square with Complex coefficients:

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

Hence zeros $x = - 1 \pm i$ $x=-1+-i$

Science

Math

Humanities

... and beyond

How do you find all the roots of $f (x) = x^{4} + 2 x^{3} + x^{2} - 2 x - 2$ $f(x) = x^4 + 2x^3 + x^2 - 2x - 2$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find all the roots of f(x) = x^4 + 2x^3 + x^2 - 2x - 2f(x)=x4+2x3+x2−2x−2f(x) = x^4 + 2x^3 + x^2 - 2x - 2?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find all the roots of $f (x) = x^{4} + 2 x^{3} + x^{2} - 2 x - 2$ $f(x) = x^4 + 2x^3 + x^2 - 2x - 2$ ?