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Diano4ka-milaya [45]
2 years ago
5

Factor each completely.one zero has been given

Mathematics
1 answer:
Vikentia [17]2 years ago
7 0

The polynomials in <em>factored</em> form are listed below:

  1. <em>f(x) = (x - 3.043) · (x - 1.521 - i 1.716) · (x - 1.521 + i 1.716)</em>
  2. <em>f(x) = (x - 2.236) · (x - 2) · (x + 2.236) · (x + 4)</em>
  3. <em>f(x) = (x + 3) · (x - 5) · (x - 1)</em>
  4. <em>f(x) = (x - 4) · (x - 3) · (x - 1) · x</em>
<h3>How to factor polynomials</h3>

In this question we must factor polynomials as there are both <em>numerical</em> and <em>analytical</em> methods. Mathematically speaking, factoring polynomials is represented by the following formula:

\sum^{n}_{i = 0} \,c_{i}\cdot x^{i} = \prod^{n}_{i=1}(x-r_{i}), \forall\,x, r_{i}\in \mathbb{C}   (1)

Where:

  • c_{i} - i-th Coefficient
  • r_{i} - i-th Root

Regarding <em>fourth order</em> polynomials we can solve them by Ferrari's method and <em>third order</em> polynomials by Descartes' method. Then, the solutions of each polynomials are given below:

<h3>Polynomial 1 (f(x) = x^{3}-4\cdot x +16)</h3>

<em>f(x) = (x - 4) · (x - 3) · (x - 1) · x</em>

<h3>Polynomial 2 (f(x) = x^{4}+2\cdot x^{3}-13\cdot x^{2}-10\cdot x + 40)</h3>

<em>f(x) = (x - 2.236) · (x - 2) · (x + 2.236) · (x + 4)</em>

<h3>Polynomial 3 (f(x) = x^{3}-3\cdot x^{2}-13\cdot x +15)</h3>

<em>f(x) = (x + 3) · (x - 5) · (x - 1)</em>

<h3>Polynomial 4 (f(x) = x^{4}-8\cdot x^{3}+19\cdot x^{2}-12\cdot x)</h3>

<em>f(x) = (x - 4) · (x - 3) · (x - 1) · x</em>

To learn more on polynomials, we kindly invite to check this verified question: brainly.com/question/17822016

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