How do you find all the zeros of $P (X) = x^{4} - 4 x^{3} - 35 x^{2} + 6 x + 144$ $P(X) = x^4 - 4x^3 - 35x^2 + 6x + 144$ ?

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How do you find all the zeros of $P (X) = x^{4} - 4 x^{3} - 35 x^{2} + 6 x + 144$ $P(X) = x^4 - 4x^3 - 35x^2 + 6x + 144$ ?

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Answer 1 · 2016-02-26T10:15:49

Zeros of $x^{4} - 4 x^{3} - 35 x^{2} + 6 x + 144$ $x^4−4x^3−35x^2+6x+144$ are $- 3$ $-3$ , $2$ $2$ and $8$ $8$

Explanation:

To find all the zeros of $P (X) = x^{4} - 4 x^{3} - 35 x^{2} + 6 x + 144$ $P(X)=x^4−4x^3−35x^2+6x+144$ , first find a factor of the independent term $144$ $144$ such as

${1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 8, -8, ....}$ for which $P(x)=0$ .

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

Hence, $(x-2)(x+3)$ or $x^2+x-6$ divides $P(x)$ . Dividing latter by former, we get

$(x^4−4x^3−35x^2+6x+144)/(x^2+x-6)$ = $x^2-5x-24$

As the RHS can be factorized to $(x-8)(x+3)$ , $8$ and $-3$ are other two zeros of $x^4−4x^3−35x^2+6x+144$ . But $-3$ has come earlier.

Hence zeros of $x^4−4x^3−35x^2+6x+144$ are $-3$ , $2$ and $8$ .

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How do you find all the zeros of $P (X) = x^{4} - 4 x^{3} - 35 x^{2} + 6 x + 144$ $P(X) = x^4 - 4x^3 - 35x^2 + 6x + 144$ ?

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Explanation:

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How do you find all the zeros of P(X) = x^4 - 4x^3 - 35x^2 + 6x + 144P(X)=x4−4x3−35x2+6x+144P(X) = x^4 - 4x^3 - 35x^2 + 6x + 144?

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Explanation:

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How do you find all the zeros of $P (X) = x^{4} - 4 x^{3} - 35 x^{2} + 6 x + 144$ $P(X) = x^4 - 4x^3 - 35x^2 + 6x + 144$ ?