How do you find all the zeros of $f (x) = x^{4} + 2 x^{3} + 22 x^{2} + 50 x - 75$ $f(x) = x^4 + 2x^3 + 22x^2 + 50x - 75$ ?

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How do you find all the zeros of $f (x) = x^{4} + 2 x^{3} + 22 x^{2} + 50 x - 75$ $f(x) = x^4 + 2x^3 + 22x^2 + 50x - 75$ ?

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Answer 1 · 2016-02-26T10:25:47

Zeros of $x^{4} + 2 x^{3} + 22 x^{2} + 50 x - 75$ $x^4+2x^3+22x^2+50x-75$ are $- 3$ $-3$ and $1$ $1$

Explanation:

To find all the zeros of $P (X) = x^{4} + 2 x^{3} + 22 x^{2} + 50 x - 75$ $P(X)=x^4+2x^3+22x^2+50x-75$ , first find a factor of the independent term $75$ $75$ such as

${1, -1, 3, -3, 5, -5, 15, -15, ....}$ for which $P(x)=0$ .

It is apparent that for $x=1$ as well as for $x=-3$ , $P(x)=0$ . Hence $1$ and $-3$ are two zeros of $x^4−4x^3−35x^2+6x+144$

Hence, $(x-1)(x+3)$ or $x^2+2x-3$ divides $P(x)$ . Dividing latter by former, we get

$(x^4+2x^3+22x^2+50x-75)/(x^2+2x-3)$ = $x^2+25$

As the RHS cannot be factorized in real factors there are no further zeros.

Hence zeros of $x^4+2x^3+22x^2+50x-75$ are $-3$ and $1$

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How do you find all the zeros of $f (x) = x^{4} + 2 x^{3} + 22 x^{2} + 50 x - 75$ $f(x) = x^4 + 2x^3 + 22x^2 + 50x - 75$ ?

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Explanation:

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How do you find all the zeros of f(x) = x^4 + 2x^3 + 22x^2 + 50x - 75f(x)=x4+2x3+22x2+50x−75f(x) = x^4 + 2x^3 + 22x^2 + 50x - 75?

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Explanation:

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How do you find all the zeros of $f (x) = x^{4} + 2 x^{3} + 22 x^{2} + 50 x - 75$ $f(x) = x^4 + 2x^3 + 22x^2 + 50x - 75$ ?