How do you find all the zeros of $f (x) = x^{4} + x^{3} + 2 x^{2} + 4 x - 8$ $f(x)= x^4+x^3+2x^2+4x-8$ with its multiplicities?

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How do you find all the zeros of $f (x) = x^{4} + x^{3} + 2 x^{2} + 4 x - 8$ $f(x)= x^4+x^3+2x^2+4x-8$ with its multiplicities?

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Answer 1 · 2016-08-03T20:35:37

$f (x)$ $f(x)$ has zeros $1$ $1$ , $- 2$ $-2$ and $\pm 2 i$ $+-2i$

Explanation:

$f (x) = x^{4} + x^{3} + 2 x^{2} + 4 x - 8$ $f(x) = x^4+x^3+2x^2+4x-8$

First note that the sum of the coefficients of $f (x)$ $f(x)$ is zero. That is:

$1 + 1 + 2 + 4 - 8 = 0$ $1+1+2+4-8 = 0$

Hence $f (1) = 0$ $f(1) = 0$ , $x = 1$ $x=1$ is a zero of $f (x)$ $f(x)$ and $(x - 1)$ $(x-1)$ is a factor:

$x^{4} + x^{3} + 2 x^{2} + 4 x - 8 = (x - 1) (x^{3} + 2 x^{2} + 4 x + 8)$ $x^4+x^3+2x^2+4x-8 = (x-1)(x^3+2x^2+4x+8)$

The remaining cubic is very regular and can be factored in a number of different ways. Let's factor it by grouping:

$x^{3} + 2 x^{2} + 4 x + 8$ $x^3+2x^2+4x+8$

$= (x^{3} + 2 x^{2}) + (4 x + 8)$ $=(x^3+2x^2)+(4x+8)$

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

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

So $x = - 2$ $x=-2$ is a zero.

The remaining quadratic has no Real zeros since $x^{2} + 4 \geq 4 > 0$ $x^2+4 >= 4 > 0$ for any Real value of $x$ $x$ , but it does have Complex zeros:

$x = \pm \sqrt{- 4} = \pm 2 i$ $x = +-sqrt(-4) = +-2i$

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How do you find all the zeros of $f (x) = x^{4} + x^{3} + 2 x^{2} + 4 x - 8$ $f(x)= x^4+x^3+2x^2+4x-8$ with its multiplicities?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find all the zeros of f(x)=x4+x3+2x2+4x−8f(x)= x^4+x^3+2x^2+4x-8 with its multiplicities?

1 Answer

Explanation:

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Impact of this question

How do you find all the zeros of $f (x) = x^{4} + x^{3} + 2 x^{2} + 4 x - 8$ $f(x)= x^4+x^3+2x^2+4x-8$ with its multiplicities?